Sruthi Reddy

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Big Ideas Math Book 8th Grade Answer Key Chapter 2 Transformations

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Performance

• Transformations STEAM Video/Performance
• Transformations Getting Ready for Chapter 2

Lesson: 1 Translations

• Lesson 2.1 Translations
• Translations Homework & Practice 2.1

Lesson: 2 Reflections

• Lesson 2.2 Reflections
• Reflections Homework & Practice 2.2

Lesson: 3 Rotations

• Lesson 2.3 Rotations
• Rotations Homework & Practice 2.3

Lesson: 4 Congruent Figures

• Lesson 2.4 Congruent Figures
• Congruent Figures Homework & Practice 2.4

Lesson: 5 Dilations

• Lesson 2.5 Dilations
• Dilations Homework & Practice 2.5

Lesson: 6 Similar Figures

• Lesson 2.6 Similar Figures
• Similar Figures Homework & Practice 2.6

Lesson: 7 Perimeters and Areas of Similar Figures

• Lesson 2.7 Perimeters and Areas of Similar Figures
• Perimeters and Areas of Similar Figures Homework & Practice 2.7

Chapter 2: Transformations 

• Transformations Connecting Concepts
• Transformations Practice Test
• Transformations Cumulative Practice

Transformations STEAM Video/Performance

STEAM Video

Shadow Puppets

Some puppets are controlled using strings or wires. How else can a puppet be controlled?

Watch the STEAM Video “Shadow Puppets.” Then answer the following questions.

Question 1.
Tory and Robert are using a light source to display puppets on a screen. Tory wants to show the pig jumping from the floor to the window. Should she use a translation, reflection, rotation, or dilation? Explain.

Answer:
In the situation given, if we translate first, we move the pre-image closer to the center of dilation than if we translate second. That will result in a different image.

Question 2.
How can Tory show the pig getting smaller as it jumps out the window?

Performance Task

Master Puppeteer

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given the coordinates of a kite being used bya puppeteer.


You will be asked to identify transformations for given movements of the kite. When might a puppeteer want to use a reflection?

Transformations Getting Ready for Chapter 2

Getting Ready for Chapter 2

Chapter Exploration

Question 1.
Work with a partner. Form each triangle on a geoboard.

• Which of the triangles are congruent to the triangle at the right?
• Measure the sides of each triangle with a ruler. Record your results in a table.
• Write a conclusion about the side lengths of triangles that are congruent.

Answer: option d

When two triangles are congruent they will have exactly the same three sides and exactly the same three angles. The equal sides and angles may not be in the same position

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what the terms might mean and record your thoughts.

Lesson 2.1 Translations

EXPLORATION 1
Work with a partner.
a. For each figure below, draw the figurein a coordinate plane. Then copy the figureonto a piece of transparent paper and slide the copy to a new location in the coordinate plane. Describe the location of the copy compared to the location of the original.

• point
• triangle
• line segment
• rectangle
• line

b. When you slide figures, what do you notice about sides, angles, and parallel lines?
c. Describe the location of each point below compared to the point A(x, y).
B(x + 1, y + 2)
C(x – 3, y + 4)
D(x – 2, y + 3)
E(x + 4, y – 1)
d. You copy a point with coordinates (x, y) and slide it horizontally a units and vertically b units. What are the coordinates of the copy?

2.1 Lesson

Try It

Tell whether the blue figure is a translation of the red figure.

Question 1.

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 2.

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same and also the size is the same. The red figure slide to form exactly blue figure. So blue figure is the translation of red figure.

Try It

Question 3.
WHAT IF?
The red triangle is translated 4 units left and 2 units up. What are the coordinates of the image?

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-2,1) B(2,5), C(1,2) and a = -4, b = 2
A'(-2+a, -2+b) = A'(-2-4, 1+2) = A'(-6,3)
B'(2+a, 5+b) = B'(2-4, 5+2) = B'(-2,7)
A'(1+a, 2+b) = C'(1-4, 2+2) = C'(-3,4)
Hence the coordinate of image are A'(-6,3), B'(-2,7), C'(-3,4).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 4.

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same but the size is different. The red figure slide to form a blue figure but it is not the same size. So blue figure is not the translation of red figure.

Question 5.

Answer:
Here in the given figure, we can see that the shape of both red and blue figures is the same, and also the size is the same. The red figure slide to form the exactly blue figure. So blue figure is the translation of red figure.

Question 6.
The vertices of a triangle are A(2, 2), B (0, 2), and C (3, 0). Translate the triangle 1 unit left and 2 units up. What are the coordinates of the image?

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(2,2) B(0,2), C(3,0) and a = -1, b = 2
A'(2+a, 2+b) = A'(2-1, 2+2) = A'(1,4)
B'(0+a, 2+b) = B'(0-1, 2+2) = B'(-1,4)
A'(3+a, 0+b) = C'(3-1, 0+2) = C'(2,2)
Hence the coordinate of image are A'(1,4), B'(-1,4), C'(2,2).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
A neighborhood planner uses a coordinate plane to design a new neighborhood. The coordinates A(1, -1), B(1, -2), and C (2, -1) represent House A, House B, and House C. The planner decides to place a playground centered at the origin, and moves the houses to make space. House A is now located at A'(3, -4). What are the new coordinates of House B and House C when each house is moved using the same translation? Justify your answer.

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(1,-1) B(1,-2), C(2,1) and A'(3,-4)
A'(1+a, -1+b) = A'(3,-4) So, a = 2, b = -3
New coordinates of the houses are
B'(1+a, -2+b) = B'(1+2, -2-3) = B'(3,-5)
C'(2+a, -1+b) = C'(2+2, -1-3) = C'(4,-4)
Hence the coordinate of image are B'(3,-5),C'(4,-4)

Question 8.
The locations of a quarterback and a wide receiver on a football field are represented in a coordinate plane. The quarterback throws the football to the point (6, -2). Use a translation to describe a path the wide receiver can take to catch the pass.

Answer:
Coordinate of Receiver: (1,3) and football point:(-6,-2)
Horizontal shift: a = x2-x1 = 6 – 1 = 5
Verrical shift: b = y2 – y1 = -2 – 3 = -5
Hence the path which receiver will take 5 unit right and 5 unit down.

Translations Homework & Practice 2.1

Review & Refresh

Solve the equation for y.

Question 1.
6x + y = 12

Answer:
Given
6x + y = 12
Subtract 6x from both sides
y = 12 – 6x
Now arranging the terms
y = -6x + 12
y = 6 (-x + 2)
Thus y = 6(-x + 2)

Question 2.
9 = x + 3y

Answer:
Given,
9 = x + 3y
3y = 9 – x
Dividing by 3 on both sides
y = (9 – x)/3
y =  \(\frac{9}{3}\) – \(\frac{x}{3}\)
Now arranging the terms
y = – \(\frac{x}{3}\) + 3
Thus y = – \(\frac{x}{3}\) + 3

Question 3.
\(\frac{1}{3}\)x + 2y = 8

Answer:
Given,
\(\frac{1}{3}\)x + 2y = 8
Subtracting x/3 from both sides
2y = 8 – \(\frac{x}{3}\)
Now arranging the terms
2y = – \(\frac{x}{3}\) + 8
y = – \(\frac{x}{6}\) + 4

Question 4.
You put $550 in an account that earns 4.4% simple interest per year. How much interest do you earn in 6 months?
A. $1.21
B. $12.10
C. $121.00
D. $145.20

Answer: $12.10

Explanation:
Given:
You put $550 in an account that earns 4.4% simple interest per year.
Principal amount: P = $550
Rate of Interest: r = 4.4%
Time: t = 6 months = 0.5 year
We know that formula for Simple Interest is SI = prt/100
SI = (550 × 4.4 × 0.5)/100
SI = 1210/100
SI = 12.10
Hence the simple interest is $12.10
Thus the correct answer is option B.

Concepts, Skills, & Problem Solving

DESCRIBING RELATIONSHIPS
For each figure, describe the location of the blue figurerelative to the location of the red figure. (See Exploration 1, p. 43.)

Question 5.

Answer: The path which the receiver will take is 6 units right and 3 units down.

Explanation:
Coordinate of Point A: (-3,2) and Point A’: (3,-5)
Horizontal shift: a = x2 – x1 = 3 – (-3) = 3 + 3 = 6
Vertical shift: b = y2 – y1 = -5 – (-2) = -5 + 2 = -3
Hence, The path which the receiver will take is 6 units right and 3 units down.

Question 6.

Answer: The path which receive will take is 5 units left and 2 units down.

Explanation:
Coordinate of point A: (3,-2) and point A’: (-2,-4)
Horizontal shift: a = x2 – x1 = -2 – (3) = -2 – 3 = -5
Vertical shift: b = y2 – y1 = -4 – (-2) = -4 + 2 = -2
Hence, The path which receive will take is 5 unit left and 2 unit down.

IDENTIFYING A TRANSLATION
Tell whether the blue figure is a translation of the red figure.

Question 7.

Answer:
Blue figure is the translation of red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure sides to form the blue figure. So, the blue figure slides to form the red figure.

Question 8.

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure.

Question 9.

Answer:
Blue figure is not the translation of the red figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figure are the same. But the orientation of the blue figure is different from the red figure. This means that the blue figure is not the translation of the red figure. The blue figure is the mirror image of the red figure.

Question 10.

Answer:
Blue figure is the translation of red-figure.

Here in the given figure, we can see that the shape of both blue and red-figure are the same, and also the size of both the figure are the same. Also, the orientation of the blue figure is the same as the red figure. This means that the red figure slides to form the blue figure. so the figure slides to form the red figure.

Question 11.

Answer:
Blue figure is the translation of red-figure.

Explanation:
Here in the given figure, we can see that the shape of both blue and red figures are the same, and also the size of both the figures are the same. Also, the orientation of the both figure is the same as the red figure. This means that the red figure slides to form the blue figure. So blue figure slides to form the red figure.

Question 12.

Answer: Blue figure is not the translation of the red-figure.

Explanation:
Here in the given figure we can see that the shape of both blue and red figure are same and the size of both the figure are not same. The red figure are smaller as compared to the blue figure. This means that blue figure is not the translation of red figure.

TRANSLATE A FIGURE
The vertices of a triangle are L(0, 1), M(1, -2), and N(-2, 1). Draw the figure and its image after the translation.

Question 13.
1 unit left and 6 units up

Answer:

explanation:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to the x-coordinate and ‘b’ is added to the y-coordinates of the vertices.
A(x,y) – A'(x+a, y+b)
The value ‘a’ and ‘b’ will be positive if the shift is Right and Vertical up and the value of ‘a’ and ‘b’ will be negative if the shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)anda=-1,b=6
L'(0+a,1+b)=l'(0-1,1+6)=L'(-1,7)
M'(1+a,-2+b)=M'(1-1,1+6)=M'(0,4)
N'(-2+a,1+b)=N'(-3-1,1+6)=N'(-4,7)
Hence,the coordinate of image are L'(-1,7),M'(0,4),N'(-4,7)

Question 14.
5 units right

Answer:
We know that to translate a figure ‘a’ units horizontal and ‘b’ units vertically in the coordinate plane, ‘a’ is added to X-coordinate and ‘b’ is added to Y-coordinate of the vertices.
a(x,y),=A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if the shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given:L(0,1),M(1,-2),N(-2,1)and a=5,b=0
L(0+a,1+b)=L'(0+5,1+0)=L'(5,1)
M'(1+a,-2=b)=M'(1+5,-2+0)=M'(6,-2)
N'(-2+a,1+b)=N'(-2+5,1+0)=N'(3,1)
Hence the coordinate of image are L'(5,1),M'(6,-2),N'(3,1).

Question 15.
(x + 2, y + 3)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x+2,y+3)
So the value of: a = 2 and b = 3
L'(o + a, 1 + b) = L'(0 + 2,1 + 3) = L'(2,4)
M'(1+a, -2 + b) = M'(1 + 2, -2 + 3) = M'(3, 1)
N'(-2 + a, 1 + b) = N'(-2 + 2, 1 + 3) = N'(0, 4)
Hence the coordinate of the image is L'(2,4), M'(3,1), N'(0,4)

Question 16.
(x – 3, y – 4)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: L(0,1), M(1,-2), N(-2,1) and (x-3, y-4)
So the value of a = -3 and b = -4
L'(0 + a, 1 + b) = L'(0-3,1-4) = L'(-3,-3)
M'(1 + a, -2 + b) = M'(1 – 3, -2-4) = M'(-2,-6)
N'(-2 + a, 1 + b) = N'(-2 – 3, 1 – 4) = N'(-5, -3)
Hence the coordinate of the image are L'(-3,-3), M'(-2,-6), N'(-5, -3)

Question 17.
YOU BE THE TEACHER
Your friend translates point A 2 units down and 1 unit right. Is your friend correct? Explain your reasoning.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points,
A(3, 1) and a = 1, b = -2
A'(3+a, 1+b) = A'(3+1, 1-2) = A'(4, -1)
So, the point A’ translated by my friend is wrong. He has reversed the x and y coordinate for translation.
Hence the correct translate point is A'(4,-1)

Question 18.
TRANSLATING A FIGURE
Translate the triangle 4 units right and 3 units down. What are the coordinates of the image?

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given points from graph: J(-1,3), K(-1,1), L(-4,1)
and a = 4, b = -3
J'(-1 + a, 3 + b) = J'(-1+4,3-3) = J'(3,0)
K'(-1 + a, 1 + b) = K'(-1 + 4, 1 – 3) = K'(3,-2)
L'(-4 + a, 1 + b) = L'(-4 + 4, 1 – 3) = L'(0,-2)
Hence the coordinate of image is J'(3,0), K'(3,-2), L'(0,-2)

Question 19.
TRANSLATING A FIGURE
Translate the figure 2 units left and 4 units down. What are the coordinates of the image?

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(-1,4), B(2,3), C(3,0), D(-1,-1)
and a = -2, b = -4
A'(-1+a, 4+b) = A'(-1-2, 4-4) = A'(-3,0)
B'(2+a, 3+b) = B'(2-2, 3-4) = B'(0,-1)
C'(3+a, 0+b) = C'(3-2, 0-4) = C'(1,-4)
D'(-1+a, -1+b) = D'(-1-2, -1-4) = D'(-3,-5)
The coordinate of image are A'(-3,0), B'(0,-1), C'(1,-4), D'(-3,-5)

DESCRIBING A TRANSLATION
Describe the translation of the point to its image.

Question 20.
(3, 2) → (1,0)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(3,-2) and the image point A’:(1,0)
Horizontal shift: a = x2 – x1 = 1 – 3 = -2
Vertical shift: b = y2 – y1 = 0 – (-2) = 0 + 2 = 2
Hence the translation path will be 2 units left and 2 units up.

Question 21.
(-8, -4) → (-3, 5)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Coordinate of the point A:(-8,-4) and the image point A’:(-3,5)
Horizontal shift: a = x2 – x1 = -3 – (-8) = 5
Vertical shift: b = y2 – y1 = 5 – (-4) = 9
Hence the translation path will be 5 units left and 9 units up.

Question 22.
REASONING
You can click and drag an icon on a computer’s desktop. Is this an example of a translation? Explain.

Answer:
Yes, the dragging of an icon on a computer’s desktop is an example of translation.
Because when dragging an icon on desktop the icon directly slides and is stored in its new position. While dragging the icon there is no change in shape and size of the icon, thus fulfilling the criteria of translation.

Question 23.
MODELING REAL LIFE
The proposed location for a new oil platform is represented in a coordinate plane by a rectangle with vertices A(1, 3), B(1, 4), C(4, 4), and D(4, -3). An inspector recommends moving the oil platform 4 units right and 2 units down. Find the coordinates of the image. Then draw the original figureand the image in the coordinate plane.

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(1,-3), B(1,4), C(4,4), D(4,-3) and a = 4, b = -2
A'(1+a, -3+b) = A'(1+4, -3-2) = A'(5,-5)
B'(1+a, 4+b) = B'(1+4, 4-2) = B'(5,2)
C'(4+a, 4+b) = C'(4+4, 4-2) = C'(8,2)
D'(4+a, -3+b) = D'(4+4, -3-2) = D'(8,-5)
Hence the coordinate of image are A'(5,-5), B'(5,2), C'(8,2), D'(8,-5)

Question 24.
PROBLEM SOLVING
A school of fish translates from point F to point D.
a. Describe the translation of the school of fish.
b. Can the fishing boat make the same translation? Explain.
c. Describe a translation the fishing boat could make to get to point D.

Answer:
a. Coordinate of the point F: (-3,2) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-3) = 2 + 5 = 7
Vertical shift: b = y2 – y1 = 3 – 2 = 1
Hence the path of translation is 5 unit Right and 1 unit Up.
b. No, the fishing boat (point B) cannot make the same translation as by fish (point F). Because in path between from point B to point D the is an island which will interrupt the translation of fishing boat.
c. Coordinate of the point B:(-2,-1) and the point D: (2,3)
Horizontal shift: a = x2 – x1 = 2 – (-2) = 2 + 2 = 4
Verical shift: b = y2 – y1 = 3 -(-1) = 3 + 1 = 4
Hence the path of translation is 4 unit Right and 4 unit Up.

Question 25.
REASONING
The vertices of a triangle are A(0, -3), B(2, -1), and C(3, -3). You translate the triangle 5 units right and 2 units down. Then you translate the image 3 units left and 8 units down. Is the original triangle identical to the final image? Explain your reasoning.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
the value of a and b will be positive if the shift is Right and vertical Up and the value of a and b will be negative if the shift is left and vertical down.
Given: A(0,-3), B(2,-1), C(3,-3) and a1 = 5, b1 = -2
A'(0+a, -3+b) = A'(0+5, -3-2) = A'(5,-5)
B'(2+a, -1+b) = B'(2+5, -1-2) = B'(7,-3)
C'(3+a, -3+b) = C'(3+5,-3-2) = C'(8,-5)
Hence the coordinate of the first image are A'(5,-5), B'(7,-3), C'(8,-5)
Given: A'(5,-5), B'(7,-3), C'(8,-5) and a2 = -3, b2 = -8
A”(5+a, -5+b) = A”(5-3, -5-8) = A”(2,-13)
B”(7+a, -3+b) = B”(7-3, -3-8) = B”(4,-11)
C”(8+a, -5+b) = C”(8-3,-5-8) = C”(5,-13)
Hence the coordinate of the first image are A”(2,-13), B”(4,-11), C”(5,-13)
a = a1+a2 = 5 – 3 = 2, and b = b1 + b2 = -2 – 8 = -10
A'(0+a, -3+b) = A'(0+2, -3-10) = A'(2,-13)
B'(2+a, -1+b) = B'(2+2, -1-10) = B'(4,-11)
C'(3+a, -3+b) = C'(3+2,-3-10) = C'(5,-13)
Hence the original triangle is identical to the final image. This is because we can use both the translation by finding the resultant translation.
For final translation we can use:(x+2, y-10)

Question 26.
DIG DEEPER!
In chess, a knight can move only in an L-shaped pattern:

• two vertical squares, then one horizontal square;
• two horizontal squares, then one vertical square;
• one vertical square, then two horizontal squares; or
• one horizontal square, then two vertical squares.

Write a series of translations to move the knight from g8 to g5.

Answer:
The series of translation to move knight from g8 to g5
1. Move 1 units Right to h8 and then 2 units Down to h6
2. Move 2 units Left to f6 and then 1 unit Up to f7
3. Move 2 units Down to f5 and then 1 unit Right to g5

Lesson 2.2 Reflections

Reflecting Figures

Work with a partner.
a. For each figure below, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Flip the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the line of symmetry.

• point
• triangle
• line segment
• rectangle
• line

b. When you reflect figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(4, 7) in terms of reflections.
d. A point with coordinates (x, y) is reflected in the x-axis. What are the coordinates of the image?
e. Repeat part(d) when the point is reflected in the y-axis

2.2 Lesson

Try It

Tell whether the blue figure is a reflection of the red figure.

Question 1.

Answer: Blue figure is not the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is not the mirror image of the red figure. Thus Blue figure is not the reflection of the red figure.

Question 2.

Answer: Blue figure is the reflection of the red figure

Explanation:
By seeing the above figure we can say that the blue figure is the mirror image of the red figure. If the red figure is flipped it would form the shape of the blue figure. Thus Blue figure is the reflection of the red figure

Try It

Question 3.
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3). Draw the figure and its reflection in (a) the x-axis and (b) the y-axis.

Answer:
Given,
The vertices of a rectangle are A(-4, -3), B(-4, -1), C(-1, -1), and D(-1, -3).
Reflection about the x-axis:
A(x,y) = A'(x,-y)
A(-4, -3) = A'(-4,3)
B(-4, -1) = B'(-4,1)
C(-1, -1) = C'(-1,1)
D(-1, -3) = D'(-1,3)
Reflection through x-axis:

Reflection through y-axis:
A(x,y) = A'(-x,y)
A(-4, -3) = A'(4,-3)
B(-4, -1) = B'(4,-1)
C(-1, -1) = C'(1,-1)
D(-1, -3) = D'(1,-3)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
REFLECTING A FIGURE
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4). Draw the figure and its reflection in
(a) the x-axis and
(b) the y-axis.

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
The vertices of a triangle are J(-3, -5), K(-2, 2), and L(1, -4).
Reflection about the x-axis:
J(-3, -5) = J'(-3,5)
K(-2, 2) = K'(-2,-2)
L(1, -4) = L'(1,4)

when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
J(-3, -5) = J'(3,-5)
K(-2, 2) = K'(2,2)
L(1, -4) = L'(-1,-4)

Question 5.
WHICH ONE DOESN’T BELONG?
Which transformation does not belong with the other three? Explain your reasoning.

Answer: 3rd figure is different from other figures. Because all the other three pictures are reflections of each other except the third one. The third picture is pointed in the same direction but all the other three figures are in opposite direction.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
You design a logo using the figure shown at the left. You want both the x-axis and the y-axis to be lines of reflection. Describe how to use reflections to complete the design. Then draw the logo in the coordinate plane.

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
A(-4,2), B(-2,2), C(0,0), D(-2,0)
Reflection about the x-axis:
A(-4,2) = A'(-4,-2)
B(-2,2) = B'(-2,-2)
C(0,0) = C'(0,0)
D(-2,0) = D'(-2,0)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Reflection about the y-axis:
A(-4,2) = A”(4,2)
B(-2,2) = B”(2,2)
C(0,0) = C”(0,0)
D(-2,0) = D”(2,0)
Now to complete the Logo again we have to take a reflection of the image figure about the y-axis. In this way, the logo will be symmetric about both axis.
A”(4,2) = A”‘(4,-2)
B”(2,2) = B”‘(2,-2)
C”(0,0) = C”‘(0,0)
D”(2,0) = D”‘(-2,0)

Question 7.
DIG DEEPER!
You hit the golf ball along the path shown, so that its final location is a reflection in the y-axis of its starting location.
a. Does the golf ball land in the hole? Explain.
b. Your friend tries the shot from the same starting location. He bounces the ball of the wall at the point (-0.5, 7) so that its path is a reflection. Does the golf ball land in the hole?

Answer:
a. Coordinates of the location of golf ball = (2,4)
Coordinates of location of hole = (-3,4)
Location of ball after reflection through y-axis = (2,4) = (-2,4)
But the location of the hole is (-3,4)
So the ball will not go into a hole and it will miss the hole by 1 unit.
Hence the ball will not go into the hole.
b. Yes, when the ball bounces at the point (-0.5,7) then it will land in the hole.

Reflections Homework & Practice 2.2

Review & Refresh

The vertices of a quadrilateral are P(-1, -1), Q(0, 4), R(3, 1), and S(1, -2). Draw the figure and its image after the translation.

Question 1.
7 units down

Answer:

We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 0, b = -7
P'(-1+a,-1+b) = P'(-1+0,-1-7) = P'(-1,-8)
Q'(0+a,4+b) = Q'(0+4,4-7) = Q'(4,-3)
R'(3+a, 1+b) = R'(3+0,1-7) = R'(3,-6)
S'(1+a,-2+b) = S'(1+0,-2-7) = S'(1,-9)
Thus the coordinate of the image is P'(-1,-8), Q'(4,-3), R'(3,-6), and S'(1,-9)

Question 2.
3 units left and 2 units up

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -3, b = 2
P'(-1+a,-1+b) = P'(-1-3,-1+2) = P'(-4,1)
Q'(0+a,4+b) = Q'(0-3,4+2) = Q'(-3,6)
R'(3+a, 1+b) = R'(3-2,1+2) = R'(0,3)
S'(1+a,-2+b) = S'(1-3,-2+2) = S'(-2,0)
Thus the coordinate of the image are P'(-4,1), Q'(-3,6), R'(0,3) and S'(-2,0)

Question 3.
(x + 4, y – 1)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = 4, b = -1
P'(-1+a,-1+b) = P'(-1+4,-1-1) = P'(3,-2)
Q'(0+a,4+b) = Q'(0+4,4-1) = Q'(4,3)
R'(3+a, 1+b) = R'(3+4,1-1) = R'(7,0)
S'(1+a,-2+b) = S'(1+4,-2-1) = S'(5,-3)
Thus the coordinate of the image are P'(3,-2), Q'(4,3), R'(7,0) and S'(5,-3)

Question 4.
(x – 5, y – 6)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinates of the vertices.
A(x,y) = A'(x+a,y+b)
the value ‘a’ and ‘b’ will be positive if shift is right and vertical up and the value of ‘a’ and ‘b’ will be negative if shift is left and vertical down.
Given,
P(-1,-1)
Q(0,4)
R(3,1)
S(1,-2) and a = -5, b = -6
P'(-1+a,-1+b) = P'(-1-5,-1-6) = P'(-6,-7)
Q'(0+a,4+b) = Q'(0-5,4-6) = Q'(-5,-2)
R'(3+a, 1+b) = R'(3-5,1-6) = R'(-2,-5)
S'(1+a,-2+b) = S'(1-5,-2-6) = S'(-4,-8)
Thus the coordinate of the image are P'(-6,-7), Q'(-5,-2), R'(-2,-5) and S'(-4,-8)

Tell whether the angles are complementary, supplementary or neither.

Question 5.

Answer:
108° + 82° = 190°
Thus the angle is neither supplementary nor complementary.

Question 6.

Answer: Complementary

Explanation:
43° + 47° = 90°
Two angles are called complementary when their measures add to 90 degrees.

Question 7.

Answer:
38° + 62° = 100°
Hence the given angle is neither supplementary nor complementary.

Question 8.
36 is 75% of what number?
A. 27
B. 48
C. 54
D. 63

Answer: B. 48

Explanation:
Let x be the unknown value.
75% of x = 36
75% × x = 36
75/100 × x = 36
3/4x × x = 36
3x = 36 × 4
3x = 144
x = 144/3
x = 48
Thus the correct answer is option B.

Concepts, Skills, &Problem Solving
DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(5, 3) in terms of reflections. (See Exploration 1, p. 49.)

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
P(x,y) = P'(x,-y)
We know that when a point is reflected about y-axis then x-coordinate becomes opposite.
P(x,y) = P'(-x,y)
Given: A(5,3), B(5,-3)
Hence the point A is reflected about the x-axis to get point B.

IDENTIFYING A REFLECTION
Tell whether the blue figure is a reflection of the red figure.

Question 12.

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 13.

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 14.

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 15.

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

Question 16.

Answer: Yes

Explanation:
The blue figure is the mirror image of the red figure. If the red figure were flipped it will result in the blue figure. So, the blue figure is a reflection of red figure.

Question 17.

Answer: No

Explanation:
The blue figure is not the mirror image of the red figure. If the red figure were flipped then the right of the blue and red figure should be facing each other. So, the blue figure is not a reflection of red figure.

REFLECTING FIGURES
Draw the figure and its reflection in the x-axis. Identify the coordinates of the image.

Question 18.
A(3, 2), B(4, 4), C(1, 3)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
A(3, 2), B(4, 4), C(1, 3)
Reflection about the x-axis:
A(3, 2) = A'(3,-2)
B(4, 4) = B'(4,-4)
C(1, 3) = C'(1,-3)
Thus the coordinate of the image are A'(3,-2), B'(4,-4), C'(1,-3)

Question 19.
M(-2, 1), N(0, 3), P(2, 2)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
M(-2, 1), N(0, 3), P(2, 2)
Reflection about the x-axis:
M(-2, 1) = M'(-2,-1)
N(0, 3) = N'(0,-3)
P(2, 2) = P'(2,-2)
Thus the coordinate of the image are M'(-2,-1), N'(0,-3), P'(2,-2)

Question 20.
H(2, -2), J(4, -1), K(6, -3), L(5, -4)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
H(2, -2), J(4, -1), K(6, -3), L(5, -4)
Reflection about the x-axis:
H(2, -2) = H'(-2,-1)
J(4, -1) = J'(4,1)
K(6, -3) = K'(6,3)
L(5, -4) = L'(5,4)
Thus the coordinate of the image are H'(-2,-1), J'(4,1), K'(6,3) and L'(5,4)

Question 21.
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)

Answer:
We know that when a point is reflected about the x-axis then y-coordinate becomes the opposite.
A(x,y) = A'(x,-y)
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
D(-2, -5), E(0, -1), F(2, -1), G(0, -5)
Reflection about the x-axis:
D(-2, -5) = D'(-2,5)
E(0, -1) = E'(0,1)
F(2, -1) = F'(2,1)
G(0, -5) = G'(0,5)
Thus the coordinate of the image are D'(-2,5), E'(0,1), F'(2,1), G'(0,5)

REFLECTING FIGURES
Draw the figure and its reflection in the y-axis. Identify the coordinates of the image.

Question 22.
Q(-4, 2), R(-2, 4), S(-1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
Q(-4, 2), R(-2, 4), S(-1, 1)
Reflection about the x-axis:
Q(-4, 2) = Q'(4,2)
R(-2, 4) = R'(2,4)
S(-1, 1)= S'(1,1)
Thus the coordinate of the image is Q'(4,2), R'(2,4), S'(1,1)

Question 23.
T(4, -2), U(4, 2), V(6, -2)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
T(4, -2), U(4, 2), V(6, -2)
Reflection about the y-axis
T(4,-2) = T'(-4,-2)
Y(4,2) = U'(-4,2)
V(6,-2) = V'(-6,-2)
Thus the coordinates of the figure are T'(-4,-2), U'(-4,2), V'(-6,-2)

Question 24.
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
W(2, -1), X(5, -2), Y(5, -5), Z(2, -4)
Reflection about the y-axis:
W(2,-1) = W'(-2,-1)
X(5,-2) = X'(-5,-2)
Y(5,-5) = Y'(-5,-5)
Z(2,-4) = Z'(-2,-4)
Thus the coordinates of the figure are W'(-2,-1), X'(-5,-2), Y'(-5,-5), Z'(-2,-4)

Question 25.
J(2, 2), K(7, 4), L(9, -2), M(3, -1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given,
J(2, 2), K(7, 4), L(9, -2), M(3, -1)
Reflection about the y-axis
J(2, 2) = J(-2,2)
K(7, 4) = K'(-7,4)
L(9, -2) = L'(-9,-2)
M(3, -1) = M'(-3,-1)
Thus the coordinates of the figure are J(-2,2), K'(-7,4), L'(-9,-2), M'(-3,-1)

Question 26.
REASONING
Which letters look the same when reflected in the line?

Answer:
The letters which will look the same after being reflected through horizontal line are
B, C, D, E, H, I, K, O, X

STRUCTURE
The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 27.
(2, -2) → (2, 2)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(2, -2) → A'(2, 2)
Here we can see that x-coordinate of both A & A’ is the same but the y-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about the x-axis.
Hence the point A is reflected about the x-axis to get point A’.

Question 28.
(-4, 1) → (4, 1)

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
when a point is reflected about the y-axis then the x coordinate becomes the opposite.
A(x,y) = A'(-x,y)
Given A(-4, 1) → A'(4, 1)
Here we can see that y-coordinate of both A & A’ is the same but the x-coordinate of A’ is just the opposite of A. This means that A’ is the reflection of A about y-axis.
Hence the point A is reflected about the y-axis to get point A’.

Question 29.
(-2, -5) → (4, -5)

Answer:
Given,
A(-2, -5) → A'(4, -5)
We observe that y-coordinate of both A and A’ is same but the x-coordinate of A’ is not opposite of A. This means that A’ is the reflection of A about a line x = a.
a = (x2+x1)/2 = (4-2)/2 = 2/2 = 1
Hence the point (-2,-5) is reflected about the line x = 1 to get point (4,-5)

Question 30.
(-3, -4) → (-3, 0)

Answer:
Given,
B(-3, -4) → B'(-3, 0)
We observe that x-coordinate of both B and B’ is the same but the y-coordinate of B’ is not the opposite of B. This means that A’ is the reflection of A about a line x = a.
b = (y2+y1)/2 = (0-4)/2 = -4/2 = -2
Hence the point (-3,-4) is reflected about the line y = -2 to get point (-3,0)

TRANSFORMING FIGURES
Find the coordinates of the figure after the transformations.

Question 31.
Translate the triangle 1 unit right and 5 units down. Then reflect the image in the y-axis.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
R(-4,1)
S(-4,4)
T(-2,1)
a = 1 and b = -5
R(-4,1) = R'(-4+a, 1+b) = R'(-4+1, 1-5) = R'(-3, -4)
S(-4,4) = S'(-4+a, 4+b) = S'(-4+1, 4-5) = S'(-3, -1)
T(-2,1) = T'(-2+a, 1+b) = T'(-2+1, 1-5) = T'(-1, -4)
Thus the coordinates of the image are R'(-3, -4), S'(-3, -1), T'(-1, -4)

Question 32.
Reflect the trapezoid in the x-axis. Then translate the image 2 units left and 3 units up.

Answer:
When a point is reflected about the x-axis then the y coordinate becomes the opposite.
A(x,y) = A'(x,-y)
Given, W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
Now reflection about the x-axis:
W(-2,-2) = W'(-2,2)
X(-2,1) = X'(-2,-1)
Y(2,1) = Y'(2,-1)
Z(4,-2) = Z'(4,2)
Thus the coordinates of the image: W'(-2,2), X'(-2,-1), Y'(2,-1), Z'(4,2)
Now translating the above image point:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
W(-2,-2), X(-2,1), Y(2,1), and Z(4,-2)
a = -2 and b = 3
W(-2,2) = W”(-2+a, 2+b) = W”(-2-2, 2+3) = W”(-4, 5)
X(-2,-1) = X”(-2+a, -1+b) = X”(-2-2, -1+3) = X”(-4, 2)
Y(2,-1) = Y”(2+a, -1+b) = Y”(2-2, -1+3) = Y”(0, 2)
Z(4,2) = Z”(4+a, 2+b) = Z”(4-2, 2+3) = Z”(2, 5)
Thus the coordinates of the image: W”(-4, 5), X”(-4, 2), Y”(0, 2), Z”(2, 5)

Question 33.
REASONING
In Exercises 31 and 32, is the original figure identical to the final image? Explain.

Answer: Yes, in exercises 31 and 32 the original figure is identical to the final image. Because the type of transformation used is reflection and translation. The shape and size of the image figure do not change when there is reflection or translation. The only position of the image changes in both cases when compared to the position of the original figure.

Question 34.
CRITICAL THINKING
Hold a mirror to the left side of the photo of the vehicle.
a. What word do you see in the mirror?

Answer: The word which we will see in the mirror will be AMBULANCE. Because the word is written in mirror image form on the vehicle.

b. Why do you think it is written that way on the front of the vehicle?

Answer: Ambulance

Explanation:
It is written in that way because the ambulance will be behind any vehicle then the word “AMBULANCE” will correctly appear in the Rear-view mirror of the front vehicle.

Question 35.
DIG DEEPER!
Reflect the triangle in the line y = x. How are the x- and y-coordinates of the image related to the x- and y-coordinates of the original triangle?

Answer:
When a point is reflected about the line y = x then both x and y-coordinate become opposite.
A(x, y) = A'(-x, -y)
Given,
D(-1,-3)
E(-1,1)
F(-3,1)
Reflection about the line y = x
D(-1,-3) = D'(1,-3)
E(-1,1) = E'(1,-1)
F(-3,1) = F'(3,-1)
Hence the coordinates of the image: D'(1,-3), E'(1,-1), F'(3,-1)

Lesson 2.3 Rotations

EXPLORATION 1
Work with a partner.
a. For each figurebelow, draw the figure in the coordinate plane. Then copy the axes and the figure onto a piece of transparent paper. Turn the transparent paper and align the origin and the axes with the coordinate plane. For each pair of figures, describe the angle of rotation.

• point
• triangle
• line segment
• rectangle

b. When you rotate figures, what do you notice about sides, angles, and parallel lines?
c. Describe the relationship between each point below and the point A(3, 6) in terms of rotations.
d. What are the coordinates of a point P(x, y) after a rotation 90° counterclockwise about the origin? 180°? 270°?

2.3 Lesson

Try It
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 1.

Answer: Yes blue figure is the rotation of red figure about the origin.

Explanation:
When we rotate the red figure 180 degrees clockwise or anti-clockwise about the origin we will get the same figure as the blue figure.

Question 2.

Answer: Blue figure is not the rotation of the red figure.

Explanation:
When the red figure is rotated about the origin in any direction the distance of the center point of both the red figure and the blue figure will be the same from the origin point (0,0). The distance between the center of the object and the center of rotation always remains the same.
Thus Blue figure is not the rotation of the red figure.

Try It

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 3.
J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3); 180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given, J(-4, -4), K(-4, 2), L(-1, 0), M(-2, -3)
Rotation about the origin
J(-4, -4) = J'(4,4)
K(-4, 2) = K'(4,-2)
L(-1, 0) = L'(1,0)
M(-2, -3) = M'(2,3)
Hence the coordinate of the image are J'(4,4), K'(4,-2), L'(1,0), M'(2,3)

Question 4.
P(-3, 2), Q(6, 1), R(-1, -5); 90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
P(-3, 2), Q(6, 1), R(-1, -5)
Rotation about the origin
P(-3, 2) = P'(-2,-3)
Q(6, 1) = Q'(-1,6)
R(-1, -5) = R'(5,-1)
Hence the coordinate of the image is P'(-2,-3), Q'(-1,6), R'(5,-1)

Question 5.
A(5, 3), B(4, -1), C(1, -1); 90° clockwise about the origin

Answer:
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
Given,
A(5, 3), B(4, -1), C(1, -1)
Rotation about the origin
A(5, 3) = A'(3,-5)
B(4, -1) = B'(-1,-4)
C(1, -1) = C'(-1,-1)
Hence the coordinate of the image are A'(3,-5), B'(-1,-4), C'(-1,-1)

Try It

Question 6.
The vertices of a triangle are P(-1, 2), Q(-1, 0), and R(2, 0). Rotate the triangle 180° about the origin, and then reflect it in the x-axis. What are the coordinates of the image?

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
P(-1, 2), Q(-1, 0), and R(2, 0)
Rotation about the origin
P(-1, 2) = P'(1,-2)
Q(-1, 0) = Q'(1,0)
R(2, 0) = R'(-2,0)
Hence the coordinate of the image is P'(1,-2), Q'(1,0), R'(-2,0)
Now reflecting above image point about x-axis:
When a point is reflected about the x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given,
P'(1,-2), Q'(1,0), R'(-2,0)
Rotation about the origin
P'(1,-2) = P”(1,2)
Q'(1,0) = Q”(1,0)
R'(-2,0) = R”(-2,0)
Hence the coordinate of the image are P'(1,2), Q'(1,0), R'(-2,0)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about point P. If so, give the angle and direction of rotation.

Answer:
Yes, the blue figure is the rotation of the red figure about the origin.

Explanation:
Because when we will rotate the red figure 90 degrees anti-clockwise about the origin we will get the same figure as the blue figure.
By this, we can say that the blue figure is the result of the rotation of red figure by 90 degrees in the clock or anti-clockwise direction.

Question 8.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.

Answer:
The statement which different from all other 3 statement is:
What are the coordinates of the image after a 270 degrees clockwise rotation about the origin?
Now coordinate of both the image are:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x, y) = A'(y, -x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2,4) = A'(4,-2)
B(4, 4) = B'(4,-4)
C(4, 1) = C'(1,-4)
Hence the coordinate of the image are: A'(4,-2), B'(4,-4), C'(1,-4)
Image of statement which different from all 3 statement
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
we know that when a point is rotated 90 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite
P(x, y) = P'(-y, x)
Given,
A(2, 4)
B(4, 4)
C(4, 1)
Rotating 90 degrees clockwise about the origin
A(2, 4) = A'(-4, 2)
B(4, 4) = B'(-4, 4)
C(4, 1) = C'(-1, 4)
Hence the coordinate of the image are: A'(4,-2), B'(-4,4), C'(-1,4)
Third statement “what are the coordinates of the image after a 270 degrees clockwise rotation about origin?” is different.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
You move the red game piece to the indicated location using a rotation about the origin, followed by a translation. What are the coordinates of the vertices of the game piece after the rotation? Justify your answer.

Answer:
To move the red game piece in the indicated location the game piee should be rotated 90 degrees in clockwise direction. The rotation of red game piece 90 degrees clockwise is same as the rotation of 270 degrees counterclockwise.
we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanges and x-coordinate becomes opposite.
A(x,y) = A'(y, -x)
Conner point of red game piece:
A(0,-1), B(0,0), C(1,0), D(1,1), E(-2,1), F(-2,0), G(-1,0), H(-1,-1)
Rotating 90 degrees clockwise about the origin:
A(0,-1) = A'(-1,0)
B(0,0) = B'(0,0)
C(1,0) = C'(0,-1)
D(1,1) = D'(1,-1)
E(-2,1) = E'(1,2)
F(-2,0) = F'(0,2)
G(-1,0) = G'(0,1)
H(-1,-1) = H'(-1,1)
Hence the coordinate of corner of red game piece are A'(-1,0), B'(0,0), C'(0,-1), D'(1,-1), E'(1,2), F'(0,2), G'(0,1), H'(-1,1)

Question 10.
DIG DEEPER!
Skytypingis a technique that airplanes use to write messages in the sky. The coordinate plane shows a message typed in the sky over a city, where the positive y-axis represents north. What does the message say? How can you transform the message so that it is read from north to south?

Answer: HELLO

Explanation:
The message above on the coordinate plane can be transformed from north to south by rotating the image 90 degrees anticlockwise.

Rotations Homework & Practice 2.3

Review & Refresh

Tell whether the blue figure is a reflection of the red figure.

Question 1.

Answer: Yes, the blue figure is the reflection of red figure.

Explanation:
Because the blue figure is the exact mirror image of the red figure. If the red figure will be flipped it will result in the blue figure. So, the blue figure is the reflection of red figure.

Question 2.

Answer: No, the blue figure is not the reflection of red figure.

Explanation:
Because the blue figure is not the mirror image of the red figure. If the red figure will be flipped it will not result in the blue figure. So blue figure is not reflection of red figure.

Find the circumference of the object. Use 3.14 or \(\frac{22}{7}\) for π.

Question 3.

Answer:
Given diameter of disk D = 28 cm
Circumference of the circular disk is π × D
C = π × 28
C = 22/7 × 28
C = 22 × 4
C = 88 cm
Thus the circumference is 88 cm.

Question 4.

Answer:
Given the diameter of disk D = 11.4 in
Circumference of the circular disk is π × D
C = π × 11.4
C = 22/7 × 11.4
C = 3.14 × 11.4
C = 35.796 in
Thus the circumference is 35.796 in

Question 5.

Answer:
Given diameter of disk r = 0.5 ft
Circumference of the circular disk is 2π × r
C = 2π × 0.5
C = 6.28 × 0.5
C = 3.14 ft
Thus the circumference is 3.14 ft

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point (2, 7) in terms of rotations. (See Exploration 1, p. 55.)

Question 6.
B(7, -2)

Answer:
Given,
A(7, 2) = B(7, -2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinates get interchanged and the x-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Hence the above rotation is 270 degrees counterclockwise about the origin.

Question 7.
C(-7, 2)

Answer:
Given,
A(7, 2) = C(-7, 2)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Hence the above rotation is 90 degrees counterclockwise about the origin.

Question 8.
D(-2, -7)

Answer:
Given,
A(2, 7) = C(-2, -7)
Here we can see that after rotation x and y coordinate are interchanged and the y-coordinate is opposite. And we know that when a point is rotated 270 degrees counterclockwise about origin then both coordinate gets interchanged and x-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Hence the above rotation is 180 degrees counterclockwise about the origin.

IDENTIFYING A ROTATION
Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and direction of rotation.

Question 9.

Answer: No the blue figure is not the rotation of the red figure.

Explanation:
Because if the blue triangle were the result of the rotation of the red triangle then the hypotenuse of the blue triangle should have been parallel to the x-axis. so, it is not the case of rotation.

Question 10.

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
Because if the red figure is rotated 90 degrees in a counterclockwise direction it will result in a blue figure.

Question 11.

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 180 degrees in counterclockwise or clockwise direction it will result in blue figure.

Question 12.

Answer: Yes, the blue figure is the result of the rotation of the red figure.

Explanation:
If the red figure is rotated 90 degrees in a clockwise direction it will result in blue figure.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 13.
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(2, -2), B(4, -1), C(4, -3), D(2, -4)
Rotating 90 degrees counterclockwise about the origin
A(2, -2) = A'(2,2)
B(4, -1) = B'(1,4)
C(4, -3) = C'(3,4)
D(2, -4) = D'(4,2)
Hence the coordinates of the image are A'(2,2), B'(1,4), C'(3,4), D'(4,2)

Question 14.
F(1, 2), G(3, 5), H(3, 2) 180° about the origin

Answer:
We know that when a point is rotated 180 degrees counterclockwise or clockwise direction about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Given,
F(1, 2), G(3, 5), H(3, 2)
Rotating 180 degrees about the origin
F(1, 2) = F'(-1,-2)
G(3, 5) = G'(-3,-5)
H(3, 2) = H'(-3,-2)
Hence the coordinates of the image are F'(-1,-2), G'(-3,-5), H'(-3,-2)

Question 15.
J(-4, 1), K(-2, 1), L(-4, -3)
90° clockwise about the origin

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
J(-4, 1), K(-2, 1), L(-4, -3)
Rotating 90 degrees clockwise about the origin
J(-4, 1) = J'(1,4)
K(-2, 1) = K'(1,2)
L(-4, -3) = L'(-3,4)
Hence the coordinates of the image are J'(1,4), K'(1,2), L'(-3,4)

Question 16.
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
270° clockwise about the origin

Answer:
The rotation of an object 270 degrees clockwise is equal to the rotation of 90 degrees counterclockwise.
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
P(-3, 4), Q(-1, 4), R(-2, 1), S(-4, 1)
Rotating 90 degrees clockwise about the origin
P(-3, 4) = P'(-4,-3)
Q(-1, 4) = Q'(-4,-1)
R(-2, 1) = R'(-1,-2)
S(-4, 1) = S'(-1,-4)
Hence the coordinates of the image are P'(-4,-3), Q'(-4,-1), R'(-1,-2), S'(-1,-4)

Question 17.
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
270° counterclockwise about the origin

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
W(-6, -2), X(-2, -2), Y(-2, -6), Z(-5, -6)
Rotating 90 degrees clockwise about the origin
W(-6, -2) = W'(-2,6)
X(-2, -2) = X'(-2,2)
Y(-2, -6) = Y'(-6,2)
Z(-5, -6) = Z'(-6,5)
Hence the coordinates of the image are W'(-2,6), X'(-2,2), Y'(-6,2), Z'(-6,5)

Question 18.
A(1, -1), B(5, -6), C(1, -6)
90° counterclockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given,
A(1, -1), B(5, -6), C(1, -6)
Rotating 90 degrees clockwise about the origin
A(1, -1) = A'(1,1)
B(5, -6) = B'(6,5)
C(1, -6) = C'(6,1)
Hence the coordinates of the image are A'(1,1), B'(6,5), C'(6,1)

Question 19.
YOU BE THE TEACHER
The vertices of a triangle are A(4, 4), B(1, -2), and C(-3, 0). Your friend finds the coordinates of the image after a rotation 90° clockwise about the origin. Is your friend correct? Explain your reasoning.

Answer:
We know that when a point is rotated 270 degrees counterclockwise about the origin then both coordinates gets interchanged and y-coordinate becomes opposite.
P(x, y) = P'(y, -x)
Given,
A(4, 4), B(1, -2), and C(-3, 0).
Rotating 90 degrees clockwise about the origin
A(4, 4) = A'(4,-4)
B(1, -2) = B'(-2,-1)
C(-3,0) = C'(0,3)
Hence the coordinates of the image are A'(4,-4), B'(-2,-1), C'(0,3)
By this I can say that my friend is not correct.

Question 20.
PROBLEM SOLVING
A game show contestant spins the prize wheel shown. The arrow remains in a fixed position while the wheel rotates. The wheel stops spinning, resulting in an image that is a rotation 270° clockwise about the center of the wheel. What is the result?

Answer: Free spin

Explanation:
The arrow is located at 90 degrees in the counterclockwise direction of free spin. So when the wheel is rotated 270 degrees in a clockwise direction the arrow will be on the free spin column.

PATTERN
A figure has rotational symmetry if a rotation of 180° or less produces an image that fits exactly on the original figure. Determine whether the figure has rotational symmetry. Explain your reasoning.

Question 21.

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem is rotated 120 degrees in any direction clockwise or counterclockwise then it will produce the same identical image. Since 120 degrees is less than 180 degrees so it will have rotational symmetry.

Question 22.

Answer: No the given figure does not have rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 360 degrees. Since 360 degrees is greater than 180 degrees so it will not have rotational symmetry.

Question 23.

Answer: Yes the given figure has rotational symmetry.

Explanation:
The given figure in the problem will produce the same identical image only when it is rotated 180 degrees. Since the maximum angle for rotational symmetry is 180 degrees so it will have rotational symmetry.

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 24.
R(-7, -5), S(-1, -2), T(-1, -5)
Rotate 90° counterclockwise about the origin. Then translate 3 units left and 8 units up.

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both coordinates gets interchanges and y-coordinate becomes opposite.
P(x,y) = P'(-y,x)
Given, R(-7, -5), S(-1, -2), T(-1, -5)
Rotating 90 degrees counterclockwise about the origin
R(-7,-5) = R'(5,-7)
S(-1, -2) = S'(2,-1)
T(-1, -5) = T'(5,-1)
The coordinate of the image are R'(5,-7), S'(2,-1), T'(5,-1)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given, R'(5,-7), S'(2,-1), T'(5,-1) and a = -3, b = 8
R'(5+a, -7+b) = R”(5-3, -7+8) = R”(2,1)
S'(2+a, -1+b) = R”(2-3, -1+8) = S”(-1,7)
R'(5+a, -1+b) = R”(5-3, -1+8) = T”(2,7)
The coordinate of the image are R”(2,1), S”(-1,7), T”(2,7)

Question 25.
J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1) Reflect in the x-axis, and then rotate 180° about the origin.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Given J(-4, 4), K(-3, 4), L(-1, 1), M(-4, 1)
Reflection about the x-axis:
J(-4, 4) = J'(-4,-4)
K(-3, 4) = K'(-3,-4)
L(-1, 1) = L'(-1,-1)
M(-4, 1) = M'(-4,-1)
The coordinate of the image are J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Now rotating the above image 180 degrees about the origin.
We know that when a point is reflected about x-axis then y-coordinate becomes opposite.
A(x, y) = A'(-x, -y)
J'(-4,-4), K'(-3,-4), L'(-1,-1), M'(-4,-1)
Rotating 180 degrees about the origin:
J'(-4,-4) = J”(4,4)
K'(-3,-4) = K”(3,4)
L'(-1,-1) = L”(1,1)
M'(-4,-1) = M”(4,1)
The coordinate of the image are J”(4,4), K”(3,4), L”(1,1), M”(4,1)

CRITICAL THINKING
Describe two different sequences of transformations in which the blue figure is the image of the red figure.

Question 26.

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 5 units towards the left to get the blue figure.
2. First rotate the red figure 90 degrees in a clockwise direction and then translate that image 1 unit towards the Right and 5 units Up to get the blue figure.

Question 27.

Answer:
Two different ways of translating a red figure in to blue figure:
1. First rotate the red figure 90 degrees in the counterclockwise direction and then translate that image 1 unit towards the left and 1 Down to get the blue figure.
2. First rotate the image in the x-axis and then translate that image 4 units towards the left and 2 units Up to get the blue figure.

Question 28.
REASONING
A trapezoid has vertices A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
a. Rotate the trapezoid 180° about the origin. What are the coordinates of the image?

Answer:
A(x, y) = A'(-x, -y)
Given,
A(-6, -2), B(-3, -2), C(-1, -4), and D(-6, -4).
Rotating 180 degrees about the origin:
A(-6, -2) = A'(6,2)
B(-3, -2) = B'(3,2)
C(-1, -4) = C'(1,4)
D(-6, -4) = D'(6,4)
The coordinates of the image of trapezoid vertices are A'(6,2), B'(3,2), C'(1,4), D'(6,4)

b. Describe a way to obtain the same image without using rotations.

Answer:
In the above question, we can see that the coordinates of all the vertices of the trapezoid are negative and all the coordinates of the image vertices are positive. So there is another way to get the vertices of the image.
First, reflect the trapezoid in the x-axis and then in the y-axis or first reflect the trapezoid in the y-axis and then in the x-axis.

ROTATING A FIGURE
The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 29.
D(2, 1), E(2, -2), F(-1, 4)
90° counterclockwise about vertex D

Answer:
P(x, y) = P'(-(y – b) + a, (x -a) + b)
Given,
D(2, 1), E(2, -2), F(-1, 4)
(a, b) = (2, 1)
Rotation about the point D(2,1)
D(2, 1) = D'(2,1)
E(2, -2) = E'(-(-2-1) + 2, (2 – 2) + 1) = E'(5, 1)
F(-1, 4) = F'(-(4 – 1) + 2, (-1 – 2) + 1) = F'(-1, -2)
Hence the coordinate of the image: D'(2,1), E'(5, 1), F'(-1, -2)

Question 30.
L(-4, -3), M(-1, -1), N(2, -2)
180° about vertex M

Answer:
When a point is rotated 180 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
L(-4, -3), M(-1, -1), N(2, -2)
Rotation about the point M(-1, -1):
L(-4, -3) = L'(-(-4 + 1) – 1, -(-3 + 1) – 1) = L'(2, 1)
M(-1, -1) = M'(-1, -1)
N(2, -2) = N'(-(2 + 1) – 1, -(-2 + 1) – 1) = N'(-4, 0)
Hence the coordinate of the image are L'(2, 1), M'(-1, -1), N'(-4, 0)

Question 31.
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
270° counterclockwise about vertex W

Answer:
When a point is rotated 270 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
W(-5, 0), X(-1, 4), Y(3, -1), Z(0, -4)
Rotation about the point W(-5, 0):
W(-5, 0) = W'(-5, 0)
X(-1, 4) = X'((4 – 0) – 5, -(-1 + 5) + 0) = X'(-1, -4)
Y(3, -1) = Y'((-1 – 0) – 5, -(3 + 5) + 0) = Y'(-6, -8)
Z(0, -4) = Z'((-4 – 0) – 5, -(0 + 5) + 0) = Z'(-9, -5)
Hence the coordinate of the image are W'(-5, 0), X'(-1, -4), Y'(-6, -8), Z'(-9, -5)

Question 32.
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
270° clockwise about vertex E.

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite and ‘b’ and ‘a’ are subtracted from x and y coordinate respectively.
P(x, y) = P'(-(x – a) + b, -(y – b) + a)
Given,
D(-3, -4), E(-5, 2), F(1, -1), G(3, -7)
Rotation about the point E(-5, 2):
D(-3, -4) = D'(-(-4 – 2) – 5, -(-3 + 5) + 2) = D'(1, 4)
E(-5, 2) = E'(-5, 2)
F(1, -1) = F'(-(-1 – 2) – 5, (1 + 5) + 2) = F'(-2, 8)
G(3, -7) = G'(-(-7 – 2) – 5, (3 + 5) + 2) = G'(4, 10)
Hence the coordinate of the image are D'(1, 4), E'(-5, 2), F'(-2, 8), G'(4, 10)

Question 33.
LOGIC
You want to find the treasure located on the map at . You are located at . The following transformations will lead you to the treasure, but they are not in the correct order. Find the correct order. Use each transformation exactly once.

• Rotate 180° about the origin.
• Reflect in the y-axis.
• Rotate 90° counterclockwise about the origin.
• Translate 1 unit right and 1 unit up.

Answer:
The correct order of transformation to get the treasure are:

• Rotate 180° about the origin.
• Rotate 90° counterclockwise about the origin.
• Reflect in the y-axis.
• Translate 1 unit right and 1 unit up.

Question 34.
DIG DEEPER!
You rotate a triangle 90° counterclockwise about the origin. Then you translate its image 1 unit left and 2 units down. The vertices of the final image are (-5, 0), (-2, 2), and (-2, -1). What are the vertices of the original triangle?

Answer:
When a point is rotated 90 degrees counterclockwise about a given point (a, b) then its both x and y coordinate becomes opposite.
Let the three vertices of the triangle be: (x1, y1), (x2, y2), (x3, y3)
P(x, y) = P'(-y, x)
Rotating 90 degrees counterclockwise about the origin:
A(x1, y1) = A'(-y1, x1)
B(x2, y2) = B'(-y2, x2)
C(x3, y3) = C'(-y3, x3)
Now translating the image of the vertex for the final image
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x, y) = A'(x + a, y + b)
Given,
A'(-y1, x1), B'(-y2, x2), C'(-y3, x3) and a = -1, b = -2
A'(-y1 + a, x1 + b) = A”(-y1 – 1, x1 – 2)
B'(-y2 + a, x2 + b) = B”(-y2 – 1, x2 – 2)
C'(-y3 + a, x3 + b) = C”(-y3 – 1, x3 – 2)
The given coordinate of vertex point of final image are: (-5, 0), (-2, 2) and (-2, -1)
Now comparing the coordinate of the final image
(-y1 – 1, x1 – 2) = (-5, 0) so y1 = 4 and x1 = 2
(-y2 – 1, x2 – 2) = (-2, 2) so y2 = 1 and x2 = 4
(-y3 – 1, x3 – 2) = (-2, 1) so y3 = 1 and x3 = 1
Hence the vertices of original triangle are (2, 4), (4, 1) and (1, 1)

Lesson 2.4 Congruent Figures

EXPLORATION 1

Work with a partner.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Then copy one of the figures onto a piece of transparent paper. Use transformations to try to obtain one of the figures from the other figure.

• A(-5, 1), B(-5, -4), C(-2, -4) and D(1, 4), E(1, -1), F(-2, -1)
• G(1, 2), H(2, -6), J(5, 0) and L(-1, -2), M(-2, 6), N(-5, 0)
• P(0, 0), Q(2, 2), R(4, -2) and X(0, 0), Y(3, 3), Z(6, -3)
• A(0, 4), B(3, 8), C(6, 4), D(3, 0) and
  F(-4, -3), G(-8, 0), H(-4, 3), J(0, 0)
• P(-2, 1), Q(-1, -2), R(1, -2), S(1, 1) and
  W(7, 1), X(5, -2), Y(3, -2), Z(3, 1)

b. Which pairs of figures in part(a) are identical? Explain your reasoning.
c. FigureA and FigureB are identical. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.4 Lesson

Try It

Question 1.
A triangle has vertices X(0, 4), Y(4, 4), and Z(4, 2). Is △XYZ congruent to any of the triangles in Example 1? Explain.

Answer:

After plotting the triangle XYZ on the coordinate plane we can say that the triangle XYZ is congruent to triangle PQR among all the given triangle in the figure. In fact, if triangle PQR is rotated 90 degrees clockwise of 270 degrees counterclockwise it will result in the triangle XYZ.

Try It

Question 2.
Describing a different Sequence of rigid motions between the figures.

Answer:
Different sequence of rigid motion to get the blue figure from the red figure are:
1. First rotate the red figure 90 degrees clockwise and the origin.
2. Then translate the image 4 units Right and 1 unit Up.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING CONGRUENT FIGURES
Use the coordinate plane shown.

a. Identify any congruent figures.

Answer: a. After seeing the figure we can say that triangle ABCD is congruent to triangle JKLM.

b. A rectangle has vertices W(4, 1), X(4, 2), Y(1, 2), and Z(-1, -1). Is Rectangle WXYZ congruent to any of the rectangles in the coordinate plane? Explain.

Answer:

Rectangle WXYZ is not congruent to any of the rectangles in the given figure because rectangle WXYZ is square of 3 units sides and the other rectangle in the figure does not have all the sides of 3 units.

RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 4.

Answer:
The sequence of the rigid motions from the red-figure to the blue figure:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure in the 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit left because one vertex (-1,-4) of red figure is on the negative side of the x-axis.

Question 5.

Answer:
The sequence of the rigid motions from red figure to blue figure:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure in the 1st quadrant.
2. Then we will translate the image 3 units right and 1 unit down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
In the coordinate plane at the left, each grid line represents 50 feet. Each figure represents a pasture.

a. Are the figures congruent? Use rigid motions to justify your answer.

Answer: No the blue figure and red figure are not congruent.

Explanation:
By reflecting the red figure about the y-axis and translating the image 4 units Up we will not the same blue figure. So both figure are not congruent to each other.

b. How many feet of fencing do you need to enclose each pasture?

Answer:
Given the length of each grid line = 50 feet
Total feet of fencing = 50 × total number of grid line along the boundary
For red figure fencing: 50 × 12 = 600 feet
For blue figure fencing: 50 × 12 = 600 feet

Question 7.
A home decorator uses a computer to design a floor tile. How can the decorator transform the tile as shown?

Answer:
First, rotate the given tiles about 90 degrees in the clockwise direction and then take the mirror image about the vertical axis.

Congruent Figures Homework & Practice 2.4

Review & Refresh

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 1.
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
90° counterclockwise about the origin

Answer:
When a point is rotated 90 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-y, x)
Given,
A(1, 3), B(2, 5), C(3, 5), D(2, 3)
Rotating 90 degrees counterclockwise about the origin:
A(1, 3) = A'(-3, 1)
B(2, 5) = B'(-5, 2)
C(3, 5) = C'(-5, 3)
D(2, 3) = D'(-3, 2)
Hence the coordinate of the image are A'(-3, 1), B'(-5, 2), C'(-5, 3), D'(-3, 2)

Question 2.
F(-2, 1), G(-1, 3), H(3, 1)
180° about the origin

Answer:
When a point is rotated 180 degrees about the origin then both x and y-coordinates become opposite.
A(x, y) = A'(-x, -y)
Given,
F(-2, 1), G(-1, 3), H(3, 1)
Rotating 90 degrees counterclockwise about the origin:
F(-2, 1) = F'(2,-1)
G(-1, 3) = G'(1,-3)
H(3, 1) = H'(-3,-1)
Hence the coordinate of the image are F'(2,-1), G'(1,-3), H'(-3,-1)

Factor the expression using the greatest common factor.

Question 3.
4n – 32

Answer:
4n – 32
Take 4 as a common factor.
4(n – 8)
Thus the greatest common factor is 4(n – 8)

Question 4.
3w + 66

Answer:
3w + 66
Take 3 as a common factor.
3(w + 22)
Thus the greatest common factor is 3(w + 22)

Question 5.
2y – 18

Answer:
2y – 18
Take 2 as a common factor.
2(y – 9)
Thus the greatest common factor is 2(y – 9).

Concepts, Skills, & Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether the figures are identical. (See Exploration 1, p. 63.)

Question 6.
G(0, 0), H(3, 2), J(1, -2) and L(-1, 0), M(2, 2), N(0, -3)

Answer:

After plotting the triangles GHJ and LMN we can say that the triangle LMN are bigger compared to the other triangle. Thus both the triangles are not identical.

Question 7.
A(-2, -1), B(-2, 2), C(-1, 1), D(-1, -2) and F(-2, 0), G(-1, 1), H(2, 1), J(1, 0)

Answer:

By seeing both the quadrilaterals ABCD and FGHJ we can say that they are identical.

IDENTIFYING CONGRUENT FIGURES
Identify any congruent figures in the coordinate plane.

Question 8.

Answer:
On observing the diagram in the given figure we can see that the shape and size of pentagon ABCDE and pentagon FKJHG are the same. The length of each side of both the pentagon is the same. Thus they are congruent.

Question 9.

Answer:
By seeing the above figure we can say that the shape and size of parallelogram EFGH and parallelogram BCDA are the same. The length of each side of both the parallelogram are same. Parallelogram BCDA can be obtained by rotating parallelogram EFGH 90 degrees clockwise and translating its image. Hence the parallelogram, EFGH is congruent to BCDA.

DESCRIBING A SEQUENCE OF RIGID MOTIONS
The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 10.

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 90 degrees clockwise about the origin because the given red figure is in the 2nd quadrant and the blue figure is in the 1st quadrant.
2. Then we will translate the image 1 unit left and 1 unit Down because one vertex of the red figure is at (-1, 1)

Question 11.

Answer:
The sequence of rigid motions between the red and blue figures are:
1. First we will rotate the red figure 180 degrees clockwise or anticlockwise about the origin because the given red figure is in 4th quadrant and the blue figure is in the 2nd quadrant.
2. Then we will translate the image 1 unit Right and 1 unit Down because one vertex of red figure is at (2, -2)

Question 12.
YOU BE THE TEACHER
Your friend describes a sequence of rigid motions between the figures. Is your friend correct? Explain your reasoning.

Answer:
When a point is reflected about x-axis then the y-coordinate becomes opposite.
A(x, y) = A'(x, -y)
Coordinates of red figure are A(1, -1), B(3, -1), C(4, -3), D(2, -3)
Reflection about the x-axis:
A(1, -1) = A'(1, 1)
B(3, -1) = B'(3, 1)
C(4, -3) = C'(4, 3)
D(2, -3) = D'(2, 3)
Now translating the above image point 5 unit left.
Coordinate of the vertex of blue figure are: A”(-4, 1), B”(3,1), C”(-1, 3), D”(-3,3)
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given:
A(1, 1), B(3, 1), C(4, 3), D(2, 3) and a = -5, b = 0
A”(1+a, 1+b) = A”(1-5, 1+0) = A”(-4, 1)
B”(3+a, 1+b) = B”(3-5, 1+0) = B”(-2,1)
C”(4+a,3+b) = C”(4-5, 3+0) = C”(-1, 3)
D”(2+a,3+b) = D”(2-5, 3+0) = D”(-3,3)
Hence the coordinate of image are A”(-4, 1), B”(-2,1), C”(-1, 3), D”(-3,3)
Since the coordinate of the vertex of the blue is the same in both ways.
We can say that the blue figure is obtained by the rigid motion of the red figure.

NAMING CORRESPONDING PARTS
The figures are congruent. Name the corresponding angles and the corresponding sides.

Question 13.

Answer:
Corresponding sides of the congruent figure are
AD = EH
AB = EF
BC = FG
CD = GH
Corresponding angles of the congruent figure are
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H

Question 14.

Answer:
Corresponding sides of the congruent figure are
PQ = WV
QR = VZ
RS = ZY
ST = YX
TP = XW
Corresponding angles of the congruent figure are
∠P = ∠W
∠Q = ∠V
∠R = ∠Z
∠S = ∠Y
∠T = ∠X

Question 15.
MODELING REAL LIFE
You use a computer program to transform an emoji. How can you transform the emoji as shown?

Answer:
First, take the reflection of that emoji about vertical line and then rotate that image 90 degrees clockwise to get that given emoji.

Question 16.
CRITICAL THINKING
Two figures are congruent. Are the areas of the two figures the same? the perimeters? Explain your reasoning.

Answer:

• The size of both figures should be the same.
• The shape of both the figures should be the same.
• All the corresponding angles should be the same.
• Both the area and perimeter of two congruent figures are the same.

Question 17.
DIG DEEPER!
The houses are identical.
a. What is the length of side LM?

Answer:
Length of LM = length of CD
length of CD = 32 feet
So, the length of LM is 32 feet

b.Which angle of JKLMN corresponds to ∠D?
Answer:
∠D = ∠M
Thus ∠M corresponds to ∠D

c. Side AB is congruent to side AE. What is the length of side AB? What is the perimeter of ABCDE?
.

Answer:
AE = JN
The length of JN is 20 ft
So, the length of AE = 20 ft
Perimeter of ABCDE = AB + BC + CD + DE + EA
Perimeter of ABCDE = 20 + 12 + 32 + 12 + 20 = 96 feet
Thus the Perimeter of ABCDE  is 96 feet

Question 18.
REASONING
Two constellations are represented by the figures in the coordinate plane shown. Are the figures congruent? Justify your answer.

Answer:
The above figure can be tranformed into below figure by rotating the figure 180 degrees clockwise or counterclockwise about the origin and translating the image 8 units Right and 8 units Up to get the above figure.

Lesson 2.5 Dilations

EXPLORATION 1

Work with a partner. Use geometry software.
a. Draw a polygon in the coordinate plane. Then dilate the polygon with respect to the origin. Describe the scale factor of the image.


b. Compare the image and the original polygon in part(a). What do you notice about the sides? the angles?
c. Describe the relationship between each point below and the point A(x, y) in terms of dilations.

d. What are the coordinates of a point P(x, y) after a dilation with respect to the origin by a scale factor of k?

2.5 Lesson

Try It

Tell whether the blue figure is a dilation of the red figure.

Question 1.

Answer:
No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 2.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding verrtices meeting at a point. This means that blue figure is dilation of the red figure.

Try It

Question 3.
WHAT IF?
Triangle ABC is dilated by a scale factor of 2. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A (1, 3), B (2, 3), C (2, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (1, 3) = A'(1 . 2, 3 . 2) = A'(2, 6)
B (2, 3) = B'(2 . 2, 3 . 2) = B'(4, 6)
C (2, 1) = C'(2 . 2, 1 . 2) =  C'(4, 2)
Hence the coordinates of the image are A'(2, 6), B'(4, 6),  C'(4, 2)

Try It

Question 4.
WHAT IF?
Rectangle WXYZ is dilated by a scale factor of \(\frac{1}{4}\). What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: W(-4, -6), X(-4, 8), Y(4, 8), Z(4, -6)
scale factor = 0.25
W(-4, -6) = W'(-4 × 0.25, -6 × 0.25) = W'(-1, -1.5)
X(-4, 8) = X'(-4 × 0.25, 8 × 0.25) = X'(-1, 2)
Y(4, 8) = Y'(4 × 0.25, 8 × 0.25) = Y'(1, 2)
Z(4, -6) = Z'(4 × 0.25, -6 × 0.25) = Z'(1, -1.5)
Hence the coordinates of the image are W'(-1, -1.5), X'(-1, 2), Y'(1, 2), Z'(1, -1.5)

Try It

Question 5.
WHAT IF?
Trapezoid ABCDis dilated using a scale factor of 3, and then rotated 180° about the origin. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Given points of trapezoid: A(-2, -1), B(-1,1), C(0,1), D(0,-1) scale factor = 3
Dilating the figure by scale factor of 3
A(-2, 1) = A'(-2 . 3, -1 . 3) = A'(-6, -3)
B(-1, 1) = B'(-1 . 3, 1 . 3) = B'(-3, 3)
C(0, 1) = C'(0 . 3, 1 . 3) = C'(0, 3)
D(0, -1) = D'(0 . 3, -1 . 3) = D'(0, -3)
Thus the coodrinate of the image are A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
when a point is rotated 180 degrees about the origin then both x and y coordinate becomes opposite.
P(x, y) = P'(-x, -y)
Image points: A'(-6, -3), B'(-3, 3), C'(0, 3), D'(0, -3)
Rotating 180 degrees about the origin:
A'(-6, -3) = A”(6, 3)
B'(-3, 3) = B”(3, -3)
C'(0, 3) = C”(0, -3)
D'(0, -3) = D”(0, 3)
Thus the coodrinate of the image are A”(6, 3), B”(3, -3), C”(0, -3), D”(0, 3)

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 6.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 7.

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 8.
DILATING A FIGURE
The vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4). Draw the image after a dilation with a scale factor of \(\frac{1}{4}\). Identify the type of dilation.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is scale factor
Given, vertices of a rectangle are J(4, 8), K(12, 8), L(12, 4), and M(4, 4), scale factor = 0.25
J(4, 8) = J'(4 × 0.25, 8 × 0.25) = J'(1, 2)
K(12, 8) = K'(12 × 0.25, 8 × 0.25) = K'(3, 2)
L(12, 4) = L'(12 × 0.25, 4 × 0.25) = L'(3, 1)
M(4, 4) = M'(4 × 0.25, 4 × 0.25) = M'(1, 1)
Hence the coordinates of the image are J'(1, 2), K'(3, 2), L'(3, 1), M'(1, 1)

Question 9.
VOCABULARY
How is a dilation different from other transformations?

Answer:
The difference between dilation and other transformations are

• In the case of dilate the size of the figure after the dilation either decrease or increase but the shape of the figure before and after dilation remains same. Also after the dilation the corresponding angles will be congruent.
• In case of other transformations such as Rotation, translation, reflection the shape and size of figure before and after transformation remains the same.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
A photograph is dilated to fit in a frame, so that its area after the dilation is 9 times greater than the area of the original photograph. What is the scale factor of the dilation? Explain.

Answer: The scale factor of length and breadth will be 3.

Explanation:
Given,
The area after the dilation is 9 times greater than the area of the original photograph.
Area = length × breadth
p = 3 × 3
Hence the scale factor of length and breadth will be 3.

Question 11.
DIG DEEPER!
The location of a water treatment plant is mapped using a coordinate plane, where each unit represents 1 foot. The plant has vertices (0, 0), (0, 180), (240, 180), and (240, 0). You dilate the figure with a scale factor of \(\frac{1}{3}\). What are the coordinates of the image? What do you need to change so that the image accurately represents the location of the plant? Explain your reasoning.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
a is the scaling factor
Location of water treatment plant: A(0, 0), B(0, 180), C(240, 180), D(240, 0)
scale factor = \(\frac{1}{3}\)
Dilating the figure by scale factor of \(\frac{1}{3}\)
A(0, 0) = A'(0 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = A'(0, 0)
B(0, 180) = B'(0 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = B'(0, 60)
C(240, 180) = C'(240 × \(\frac{1}{3}\), 180 × \(\frac{1}{3}\)) = C'(80, 60)
D(240, 0) = D'(240 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = D'(80, 0)
Hence the coordinates of the image are A'(0, 0), B'(0, 60), C'(80, 60), D'(80, 0)

Dilations Homework & Practice 2.5

Review & Refresh

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 1.

Answer:
Sequence of rigid motion between the red and blue figure are
1. First rotate the blue figure 90 degrees in counterclockwise direction about the orgin because blue figure in 1st quadrant and red figure is in 3rd quadrant.
2. Then translate the image 1 unit left and 4 units down.

Question 2.

Answer:
Sequence of rigid motion between the red and blue figure are
1. First reflect the blue fiure about x-axis. The image after the reflection will lies in the 3rd quadrant with same orientation.
2. Then translate the image 5 units Right.

Tell whether the ratios form a proportion.

Question 3.
3 : 5 and 15 : 20

Answer:
When two ratios are equal then it is called as proportion.
Given,
Given 3 : 5 and 15 : 20
3/5 and 15/20
3/5 and 3/4
Since the above two ratio are not equal hence they are not proportion

Question 4.
2 to 3 and 12 to 18

Answer:
When two ratios are equal then it is called as proportion.
Given,
2 to 3 and 12 to 18
2/3 and 12/18
2/3 and 2/3
Since the above two ratio are equal hence they are proportion.

Question 5.
7 : 28 and 12 : 48

Answer:
When two ratios are equal then it is called as proportion.
Given,
7 : 28 and 12 : 48
7/28 and 12/48
1/4 and 1/4
Since the above two ratio are equal hence they are proportion.

Concepts, Skills, &Problem Solving

DESCRIBING RELATIONSHIPS
Describe the relationship between the given point and the point A(8, 12) in terms of dilations. (See Exploration 1, p. 69.)

Question 6.
B(16, 24)

Answer:
Given a point and its image: A(8,12), B(16, 24)
Here we can see that both x-coordinate and y-coordinate of image point have increased to double.
This means that in this case, the image figure has become larger by the scale factor of 2 with respect to the origin.
Hence the dilation scale factor is 2.

Question 7.
C(2, 3)

Answer:
Given a point and its image: A(8,12), C(2, 3)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to one-fourth.
This means that in this case, the image figure has become smaller by the scale factor of 0.25 with respect to the origin.
Thus the dilation scale factor is 1/4.

Question 8.
D(6, 9)

Answer:
Given a point and its image: A(8,12), D(6, 9)
Here we can see that both x-coordinate and y-coordinate of image point has decreased to three-fourth.
This means that in this case the image figure has become smaller by the scale factor of 0.75 with respect to the origin.
Thus the dilation scale factor is 3/4.

IDENTIFYING A DILATION
Tell whether the blue figure is a dilation of the red figure.

Question 9.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 10.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of the red figure are congruent to the angles of the red figure. Also there will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is a dilation of the red figure.

Question 11.

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
The Blue and red figure has same size and same shape but the blue figure is a reflection of vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

Question 12.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 13.

Answer: Yes, the blue figure is a dilation of the red figure.

Explanation:
We observe that all the angles of red figure are congruent to the angles of red figure. Also their will be lines connecting the corresponding vertices meeting at a point. This means that blue figure is dilation of the red figure.

Question 14.

Answer: No the blue figure is not the dilation of the red figure.

Explanation:
Blue and red figure has same size and same shape but the blue figure is reflection about vertical axis. So, the lines corresponding vertices of both figure does not meet at a point. This means that the blue figure is not the dilation of the red figure.

DILATING A FIGURE
The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 15.
A(1, 1), B(1, 4), C(3, 1); k = 4

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(1, 1), B(1, 4), C(3, 1) and scale factor = 4
Dilating the figure by scale factor by 4
A(1, 1) = A'(1 × 4, 1 × 4) = A'(4, 4)
B(1, 4) = B'(1 × 4, 4 × 4) = B'(4, 16)
C(3, 1) = C'(3 × 4, 1 × 4) = C'(12, 4)
Hence the coordinate of the image are A'(4, 4), B'(4, 16), C'(12, 4)

Question 16.
D(0, 2), E(6, 2), F(6, 4); k = 0.5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: D(0, 2), E(6, 2), F(6, 4) and scale factor = 0.5
Dilating the figure by scale factor by 0.5
D(0, 2) = D'(0 × 0.25, 2 × 0.25) = D'(0, 1)
E(6, 2) = E'(6 × 0.25, 2 × 0.25) = E'(3, 1)
F(6, 4) = F'(6 × 0.25, 4 × 0.25) = F'(3, 2)
Hence the coordinate of the image are D'(0, 1), E'(3, 1), F'(3, 2)

Question 17.
G(-2, -2), H(-2, 6), J(2, 6); k = 0.25

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: G(-2, -2), H(-2, 6), J(2, 6) and scale factor = 0.25
G(-2, -2) = G'(-2 × 0.25, -2 × 0.25) = G'(-0.5, -0.5)
H(-2, 6) = H'(-2 × 0.25,6 × 0.25) = H'(-0.5, 1.5)
J(2, 6) = G'(2 × 0.25, 6 × 0.25) = J'(0.5, 1.5)
Hence the coordinate of the image are G'(-0.5, -0.5), H'(-0.5, 1.5), J'(0.5, 1.5)

Question 18.
M(2, 3), N(5, 3), P(5, 1); k = 3

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: M(2, 3), N(5, 3), P(5, 1) and scale factor = 3
M(2, 3) = M'(2 × 3, 3 × 3) = M'(6, 9)
N(5, 3) = N'(5 × 3, 3 × 3) = N'(15, 9)
P(5, 1) = P'(5 × 3, 1 × 3) = P'(15, 3)
Hence the coordinate of the image are M'(6, 9), N'(15, 9), P'(15, 3)

Question 19.
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0); k = \(\frac{1}{3}\)

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
Q(-3, 0), R(-3, 6), T(4, 6), U(4, 0) and scale factor = \(\frac{1}{3}\)
Q(-3, 0) = Q'(-3 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = Q'(-1, 0)
R(-3, 6) = R'(-3 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = R'(-1, 2)
T(4, 6) = T'(4 × \(\frac{1}{3}\), 6 × \(\frac{1}{3}\)) = T'(4/3, 2)
U(4, 0) = U'(4 × \(\frac{1}{3}\), 0 × \(\frac{1}{3}\)) = U'(4/3, 0)
Hence the coordinate of the image are Q'(-1, 0), R'(-1, 2), T'(4/3, 2), U'(4/3, 0)

Question 20.
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2); k = 5

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
V(-2, -2), W(-2, 3), X(5, 3), Y(5, -2), scaling factor = 5
Dilating the figure by scale factor of 5
V(-2, -2) = V'(-2 × 5, -2 × 5) = V'(-10, -10)
W(-2, 3) = W'(-2 × 5, 3 × 5) = W'(-10, 15)
X(5, 3) = X'(5 × 5, 3 × 5) = X'(25, 15)
Y(5, -2) = Y'(5 × 5, -2 × 5) = Y'(25, -10)
Hence the coordinate of the image are V'(-10, -10),W'(-10, 15), X'(25, 15), Y'(25, -10)

Question 21.
YOU BE THE TEACHER
Your friend finds the coordinates of the image of △ABC after a dilation with a scale factor of 2. Is your friend correct? Explain your reasoning.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
The points of triangle: A(2, 5), B(2, 0), C(4, 0)
scale factor = 2
Dilating the figure by scale factor of 2
A(2, 5) = A'(2 × 2, 5 × 2) = A'(4, 10)
B(2, 0) = B'(2 × 2, 0 × 2) = B'(4, 0)
C(4, 0) = C'(4 × 2, 0 × 2) = C'(8, 0)
Hence the coordinate of the image are A'(4, 10), B'(4, 0), C'(8, 0)
By this, we can say that my friend is correct.

FINDING A SCALE FACTOR
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.

Question 22.

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = A’B’/AB = 6/3 = 2
Scale factor = 2
Hence, type of dilation is enlargement with scale factor of 2.

Question 23.

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = X’Y’/XY= 2/8 = 1/4
Scale factor = 1/4
Hence, the type of dilation is reduction with the scale factor of 1/4

Question 24.

Answer:
Scale factor = side length of image/side length of the original figure
Scale factor = J’K’/JK = 15/10 = 3/2
Scale factor = 3/2
Hence, the type of dilation is reduction with the scale factor of 3/2

Question 25.

Answer:
Scale factor = side length of image/side length of original figure
Scale factor = Q’R’/QR = 4/8 = 1/2
Scale factor = 1/2
Hence, type of dilation is reduction with scale factor of 1/2

USING MORE THAN ONE TRANSFORMATION
The vertices of a figure are given. Find the coordinates of the image after the transformations given.

Question 26.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2.

Answer:
We know that when a point is reflected about y-axis then is x-coordinate becomes opposite.
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
A(x, y) = A'(-x, y)
A(-5, 3) = A'(5, 3)
B(-2, 3) = B'(2, 3)
C(-2, 1) = C'(2, 1)
D(-5, 1) = D'(5, 1)
Coordinate of the image are A'(5, 3), B'(2, 3), C'(2, 1), D'(5, 1)
A(-5, 3), B(-2, 3), C(-2, 1), D(-5, 1)
Reflect in the y-axis. Then dilate using a scale factor of 2
A'(5, 3) = A”(5 × 2, 3 × 2) = A”(10, 6)
B'(2, 3) = B”(2 × 2, 3 × 2) = B”(4, 6)
C'(2, 1) = C”(2 × 2, 1 × 2) = C”(4, 2)
D'(5, 1) = D”(5 × 2, 1 × 2) = D”(10, 2)
Coordinate of the image are A”(10, 6), B”(4, 6), C”(4, 2), D”(10, 2)

Question 27.
F(-9, -9), G(-3, -6), H(-3, -9).
Dilate using a scale factor of \(\frac{2}{3}\). Then translate 6 units up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given,
F(-9, -9), G(-3, -6), H(-3, -9) and scale factor of \(\frac{2}{3}\)
Dilating the figure by scale factor of \(\frac{2}{3}\)
F(-9, -9) = F'(-9 × 2/3, -9 × 2/3) = F'(-6, -6)
G(-3, -6) = G'(-3 × 2/3, -6 × 2/3) = G'(-2, -4)
H(-3, -9) = H'(-3 × 2/3, -9 × 2/3) = H'(-2, -6)
Coordinate of the image are F'(-6, -6), G'(-2, -4), H'(-2, -6)
Now translating above image 6 units up
F'(-6, -6), G'(-2, -4), H'(-2, -6) and a = 0, b = 6
F”(-6 + a, -6 + b) = F”(-6 + 0, -6 + 6) = F”(-6, 0)
G”(-2 + a, -4 + b) = G”(-2 + 0, -4 + 6) = G”(-2, 2)
H”(-2 + a, -6 + b) = H”(-2 + 0, -6 + 6) = H”(-2, 0)
Coordinate of the image are F”(-6, 0), G”(-2, 2), H”(-2, 0)

Question 28.
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin. Then dilate using a scale factor of 3.

Answer:
The rotation of an object 90 degrees clockwise is equal to the rotation of 270 degrees counterclockwise.
When a point is rotated 270 degrees counterclockwise about the origin then both x and y-coordinates gets interchanged and the x-coordinate becomes the opposite.
A(x, y) = A'(y, -x)
J(1, 1), K(3, 4), L(5, 1)
Rotate 90° clockwise about the origin.
J(1, 1) = J'(1, -1)
K(3, 4) = K'(4, -3)
L(5, 1) = L'(1, -5)
Coordinate of the image are J'(1, -1), K'(4, -3), L'(1, -5)
Now dilate using a scale factor of 3.
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
J'(1, -1) = J”(1 . 3, -1 . 3) = J”(3, -3)
K'(4, -3) = K”(4 . 3, -3 . 3) = K”(12, -9)
L'(1, -5) = L”(1 . 3, -5 . 3) = L”(3, -15)
Coordinate of the image are J”(3, -3), K”(12, -9), L”(3, -15)

Question 29.
LOGIC
You can use a flashlight and a shadow puppet (your hands) to project shadows on the wall.
a. Identify the type of dilation.p

Answer: The type of dilation is an enlargement

b. What does the flashlight represent?

Answer: Flashlight represents center of dilation because all the line connecting shadow and hand meet at the flashlight.
c. The length of the ears on the shadow puppet is 3 inches. The length of the ears on the shadow is 4 inches. What is the scale factor?

Answer: Scale factor = length of ears on shadow/length of ears on puppet
Scale factor = 4/3
d. Describe what happens as the shadow puppet moves closer to the flashlight. How does this affect the scale factor?

Answer:
As the flashlight will move closer the shadow will become larger. Also, the scale factor will increase.

Question 30.
REASONING
A triangle is dilated using a scale factor of 3. The image is then dilated using a scale factor of \(\frac{1}{2}\). What scale factor can you use to dilate the original triangle to obtain the final image? Explain.

Answer:
Given the first scale factor of triangle S1 = 3
Given second scale factor of triangle S2 = 1/2
We know that the final scale factor S = S1 × S2
Final Scale factor S = 3 × 1/2 = 3/2
Hence, the scale factor of the final image will be the multiplication of the first and second dilation scale factor and the final scale factor will be 3/2.

CRITICAL THINKING
The coordinate notation shows how the coordinates of a figure are related to the coordinates of its image after transformations. What are the transformations? Are the figure and its image congruent? Explain.

Question 31.
(x, y) → (2x + 4, 2y – 3)

Answer:
Given, (x, y) → (2x + 4, 2y – 3)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 2.
Also, 4 has been added to x-coordinate while 3 is added to y-coordinate which means that obtained after the dilation has been translated 4 unit Right and 3 units Down.
The final image will not be congruent because after the dilation the size of the image either increases or decreases that depend on the type of dilation.

Question 32.
(x, y) → (-x – 1, y – 2)

Answer:
Given, (x, y) → (-x – 1, y – 2)
We can see that 1 has been subtracted from x-coordinate while 2 is subtracted from y-coordinate which means that will image has translated 1 unit left and 2 units down. And also x-coordinate is opposite which means the image has been reflected about the y-axis.
Hence, transforms translation of 1 unit left and 2 units down followed by reflection about y-axis.

Question 33.

Answer:
Given, (x, y) → (1/3x, -1/3y)
We can see that both x-coordinate and y-coordinate has been multiplied by 2 this means that the point has been dilated by the scale factor of 1/3. Also, y-coordinate is opposite which means that image obtained after the dilation has been reflected about the x-axis.
Thus transforms are dilation with the scale factor of 1/3 followed by reflection about the x-axis.

STRUCTURE
The blue figure is a transformation of the red figure. Use coordinate notation to describe the transformation. Explain your reasoning.

Question 34.

Answer:
Coordinates of original figure A(1, 1) B(1, 2), C(2, 1)
Coordinates of red figure A'(2, 3) B'(2, 6), C'(4, 3)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 2/1 = 2
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 3/1 = 3
Thus to transfer the red-figure into the blue figure x-coordinate of all the points has been multiplied by 2 and the y-coordinate of all the points has been multiplied by 3.

Question 35.

Answer:
Coordinates of original figure A(4, 4) B(4, 8), C(8, 8), D(8, 4)
Coordinates of red figure A'(1, 2) B'(1, 4), C'(2, 4), D'(2. 2)
Scale factor of x-coordinate = x-coordinate of image/x-coordinate of image = 1/4 = 0.25
Scale factor of y-coordinate = y-coordinate of image/y-coordinate of image = 2/4 = 1/2 = 0.5
Thus to transfer the red figure into the blue figure x-coordinate of all the points has been multiplied by 0.25 and the y-coordinate of all the points has been multiplied by 0.50

Question 36.
NUMBER SENSE
You dilate a figure using a scale factor of 2, and then translate it 3 units right. Your friend translates the same figure 3 units right and then dilates it using a scale factor of 2. Are the images congruent? Explain.

Answer:
Blue the final image in both the case will be of the same shape and size.
Yes, the image in both cases will be the same.

Question 37.
PROBLEM SOLVING
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2). Dilate the trapezoid with respect to vertex A using a scale factor of 2. What are the coordinates of the image? Explain the method you used.

Answer:
When the points of a given figure are dilated about a point we simply multiply the distance of each side by the given scale factor. The coordinate of one point remains the same about which dilation occurs.
The vertices of a trapezoid are A(-2, 3), B(2, 3), C(5, -2), and D(-2, -2).
Scale factor = 2
So, here the coordinate of point A(-2, 3) will remains the same but all the other coordinates of points B’, C’, D’ will change according to the distance between each side of the trapezoid.

Image of the figure after dilating by a scale factor of 2
A(-2, 3) = A'(-2, 3)
B(2, 3) = B'(6, 3)
C(5, -2) = C'(12, -7)
D(-2, -2) = D'(-2, -7)
Thus the coordinate of the image are A'(-2, 3), B'(6, 3), C'(12, -7), D'(-2, -7)

Question 38.
DIG DEEPER!
A figure is dilated using a scale factor of -1. How can you obtain the image without using a dilation? Explain your reasoning.

Answer:
When a figure is dilated using a scale factor of -1 then both the x and y-coordinate of the image will become opposite.
Example:
A'(x . -1, y . -1) = A'(-x, -y)
But there are two ways to get the same image:
1. By rotating the figure 180 degrees clockwise or anticlockwise
A(x, y) rotating 180 degrees about the origin = A'(-x, -y)
2. By rotating the figure about the x-axis and y axis
A(x, y) reflecting about the origin = A'(-x, -y)

Lesson 2.6 Similar Figures

EXPLORATION 1

Work with a partner. Use geometry software.
a. For each pair of figures whose vertices are given below, draw the figures in a coordinate plane. Use dilations and rigid motions to try to obtain one of the figures from the other figure.

• A(-3, 6), B(0, -3), C(3, 6) and G(-1, 2), H(0, 1), J(1, 2)
• D(0, 0), E(3, 0), F(3, 3) and L(0, 0), M(0, 6), N(-6, 6)
• P(1, 0), Q(4, 2), R(7, 0) and X(-1, 0), Y(-4, 6), Z(-7, 0)
• A(-3, 2), B(-1, 2), C(-1, -1), D(-3, -1) and F(6, 4), G(2, 4), H(2, -2), J(6, -2)
• P(-2, 2), Q(-1, -1), R(1, -1), S(2, 2) and W(2, 8), X(3, 3), Y(7, 3), Z(8, 8)

b. Is a scale drawing represented by any of the pairs of figures in part(a)? Explain your reasoning.
c. Figure A is a scale drawing of Figure B. Do you think there must be a sequence of transformations that obtains Figure A from Figure B? Explain your reasoning.

2.6 Lesson

Try It

Question 1.
A triangle has vertices D(0, 4), E(5, 4), and F(5, 0). Is △DEF similar to △ABC and △JKL in Example 1? Explain.

Answer:
Given coordinate of the triangle ABC: A(0, 3), B(3, 3), C(3, 0)
Given coordinate of the triangle DEF: D(0, 4), E(5, 4), F(5, 0)
Given coordinate of the triangle JKL: J(0, 6), K(6, 6), L(6, 0)
Here we can see that there is no fixed relation between the coordinate between triangle ABC and DEF or triangle ABC and JKL. So no triangle is dilation with the triangle ABC.
Hence, triangle ABC is not similar △DEF and △JKL.

Try It

Question 2.
Can you reflect the red figure first, and then perform the dilation to obtain the blue figure? Explain.

Answer:
Because the final image will not depend on the order of transformation. When we will first reflect red figure then the image will be of the same size and after the dilation of the image obtained after reflection we will get the same image so we can use any two method but the final image will be the same

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 3.
IDENTIFYING SIMILAR FIGURES
In the coordinate plane at the left, determine whether Rectangle ABCD is similar to Rectangle EFGH. Explain your reasoning.

Answer: No rectangle ABCD is not similar to rectangle EFGH

Explanation:
Because the orientation of rectangle ABCD is not the same as the rectangle EFGH. Also, rectangle ABCD is not dilated with rectangle EFGH. So there is no similarity transformation between rectangle ABCD and rectangle EFGH.

Question 4.
SIMILARITY TRANSFORMATION
The red triangle is similar to the blue triangle. Describe a similarity transformation between the figures.

Answer:
Coordinate of left vertex of the red triangle: A(5, 5)
Coordinate of the same vertex after dilation: A'(10, 10)
Now coordinate left vertex of the blue triangle: A”(0, 2)
So, the value of a = 0 – 10 = -10 and b = 2 – 10 = -8
It is given that the red triangle is similar to the blue triangle so the steps of transformation:
First, dilate the figure by the scale factor of 2 and then translate the image 10 units left and 8 units down.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
A medical supplier sells gauze in large and small rectangular sheets. A large sheet has a length of 9 inches and an area of 45 square inches. A small sheet has a length of 4 inches and a width of 3 inches. Are the sheets similar? Justify your answer.

Answer:
Condition for the rectangular sheets to be similar is that all the corresponding sides of bigger and smaller rectangular sheets should be in proportional.
Area of larger rectangular sheets a = 45 sq in
length of larger rectangular sheets l1 = 9 in
Width = a/l = 45/9 = 5 in
length of smaller rectangular sheets l2 = 4 in
width of smaller rectangular sheets b2 = 3 in
condition for similarity l1/l2 = b1/b2
9/4 ≠ 5/3
These sheets are not similar

Question 6.
The sail on a souvenir boat is similar in shape to the sail on a sailboat. The sail on the sailboat is in the shape of a right triangle with a base of 9 feet and a height of 24 feet. The height of the souvenir’s sail is 3 inches. What is the base of the souvenir’s sail?

Answer:
Given,
The base of sail on sailboat b1 = 9 ft
height of sail on sailboat h1 = 24 ft
Given height of sail on souvenir boat: h2 = 3in = 0.25 ft
h1/h2 = b1/b2
24/0.25 = 9/h2
h2 = (9 × 0.25)/24 = 0.9375 ft = 1.125 in
Thus the height of sail of a souvenir boat is 1.125 in

Question 7.
DIG DEEPER!
A coordinate plane is used to represent a cheerleading formation. The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation. Three vertices of the new formation are J(-2, -2), K(0, -4), and L(2, -2). What is the location of the fourth vertex? Explain.

Answer:
The vertices of the formation are A(4, 4), B(0, 8), C(4, 4), and D(0, 6). A choreographer creates a new formation similar to the original formation.
We observe the image point carefully that both the x and y coordinate of the image point is just half of the original point and each y-coordinate is opposite.
A(4, 4) = J(-2, -2)
B(0, 8) = K(0, -4)
C(4, 4) = L(2, -2)
D(0, 6) = M(x, y)
This means that the point A, B, C are dilated by using a scale factor of 0.5 and the image obtained from the dilation is reflected about the x-axis.
So, the image point is D(0, 6) = M(0, -3)

Similar Figures Homework & Practice 2.6

Review & Refresh

Tell whether the blue figure is a dilation of the red figure.

Question 1.

Answer: No

Explanation:
Because the shape and size of both red and blue figure are the same which is not the property of dilation. The blue figure is the result of the reflection of the red figure 180 degrees in the clockwise or counterclockwise direction.

Question 2.

Answer: Yes

Explanation:
When we see both red and blue figures closely we observe that all the angles of the red figure are congruent to the blue figure. Also, there will be the lines connecting corresponding vertices meeting at a point. This means that the blue figure is a dilation of red figure.

Question 3.
You solve the equation S = lw + 2wh for w. Which equation is correct?

Answer: Option C

Explanation:
S = lw + 2wh
lw + 2wh = S
Taking w as a common factor
w(l + 2h) = s
w = s/(l + 2h)
Thus the correct answer is option C.

Concepts, Skills, &Problem Solving
TRANSFORMING FIGURES
The vertices of a pair of figures are given. Determine whether a scale drawing is represented by the pair of figures. (See Exploration 1, p. 77.)

Question 4.
A(-8, -2), B(-4, 2), C(-4, -2) and G(2, -1), H(4, -1), J(2, -3)

Answer: Yes

Explanation:

After plotting both triangles we see that the original figure is exactly double of the image. Each side of the original triangle is double the length of the image triangle. So given vertices pair represent a scale drawing.
Scale factor = 2

Question 5.
A(0, 3), B(3, 4), C(5, 3), D(3, 2) and F(-4, 4), G(-1, 5), H(5, 3), J(3, 2)

Answer: No

Explanation:

After plotting both given figures we can see that there is no transformation relation between the original figure and the image figure. So, the given vertices pair does not represent a scale drawing.

IDENTIFYING SIMILAR FIGURES
Determine whether the figures are similar. Explain your reasoning.(See Exploration 1, p. 77.)

Question 6.

Answer: Rectangle ABCD is similar to rectangle EFGH.

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of rectangle ABCD and rectangle EFGH are equal. And also the corresponding sides of both rectangle are in proportional.
∠A = ∠E, ∠B = ∠F, ∠C = ∠G, ∠D = ∠H
AB/EF = BC/FG = GH/CD = DA/HE = 1/2
Hence rectangle ABCD is similar to rectangle EFGH.

Question 7.

Answer: Both the triangle are not similar

Explanation:
Because when you see both the given figure we can see that all the corresponding angles of triangle ABC and triangle JKL are equal. And also the corresponding sides of both triangles are not proportional.
AB/JK ≠ KL/BC ≠ CA/LJ
Hence triangle ABC is not similar to triangle JKL

IDENTIFYING SIMILAR FIGURES
Draw the figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.

Question 8.
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)
Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)
Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)

Answer: Rectangle A and B are similar

Explanation:
Rectangle A: (0, 0), (4, 0), (4, 2), (0, 2)

Rectangle B: (0, 0), (6, 0), (6, 3), (0, 3)

Rectangle C: (0, 0), (4, 0), (4, 2), (0, 2)

By seeing the above figure we can say that rectangle A and rectangle B are similar and rectangle A and Rectangle C are congruent.

Question 9.
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)
Figure B: (1, 4), (4, 4), (4, 1), (1, 1)
Figure C: (2, 1), (5, 1), (5, 3), (2, 3)

Answer: Rectangle A and B are similar

Explanation:
FigureA: (4, 2), (2, 2), (2, 0), (4, 0)

Figure B: (1, 4), (4, 4), (4, 1), (1, 1)

Figure C: (2, 1), (5, 1), (5, 3), (2, 3)

Rectangles A and B are similar because in rectangle A and B all the corresponding angles are equal and also all the corresponding sides are equal.

DESCRIBING A SIMILARITY TRANSFORMATION
The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 10.

Answer:
1. First rotate the red figure 90 degrees anticlockwise because the red figure is in the first quadrant and the blue figure is in the second quadrant.
2. Then dilate the image obtained after the rotation by the scale factor of 2 because the blue figure is double the size of the red figure.
Scale factor = side of the image/side of the original figure = 4/2 = 2

Question 11.

Answer:
First, dilate the red figure by the scale figure of 3 because the blue figure is triple the size of red figure.
scale factor = side of image/side of original figure = 6/2 = 3

Question 12.
MODELING REAL LIFE
A barrier in the shape of a rectangle is used to retain oil spills. On a blueprint, a similar barrier is 9 inches long and 2 inches wide. The width of the actual barrier is 1.2 miles. What is the length of the actual barrier?

Answer:
Given,
Width of the actual barrier = 1.2 miles
Width of the barrier in the blueprint = 2 inches
2 inch dimension of blueprint = 1.2 miles of original
So, 1 inch dimension of blueprint = 1.2/2 = 0.6 miles.
Since the length of the barrier in the blueprint = 9 inches,
Thus the length of the actual barrier = 9(0.6) = 5.4 miles.

Question 13.
LOGIC
Are the following figures always, sometimes, or never similar? Explain.
a. two triangles
b. two squares
c. two rectangles

Answer:
a. Two triangles sometimes two triangles are similar when all the corresponding angles are equal and all the corresponding sides lengths are in proportion.
b. Two squares always two square are similar only when all the sides are proportional and all the angles are equal.
c. Two rectangles are similar when all the corresponding angles are equal but the lengths of the corresponding sides are not always in proportion.

Question 14.
CRITICAL THINKING
Can you draw two quadrilaterals each having two 130° angles and two 50° angles that are not similar? Justify your answer.

Answer:
Quadrilateral 1: 50°, 50°, 130°, 130° (trapezoid)
Quadrilateral 2: 50°, 130°, 50°, 130° (parallelogram)

Question 15.
REASONING
The sign is rectangular.
a. You increase each side length by 20%. Is the new sign similar to the original? Explain your reasoning.

Answer:
Given,
You increase each side length by 20%
Scale factor = 1 + percentage increase/100 = 1 + 20/100 = 1.2

b. You increase each side length by 6 inches. Is the new sign similar to the original? Explain your reasoning.

Answer:
No, because when length and width are of a different size then adding 6 inches on each side will not increase the figure in a fixed proportion. So the corresponding length will be not proportional. This means that both the figure will be not in dilation, hence not similar.

Question 16.
DIG DEEPER!
A person standing 20 feet from a streetlight casts a shadow as shown. How many times taller is the streetlight than the person? Assume the triangles are similar.

Answer:
Length of shadow l1 = 10 ft
height of man b1 = 6 ft
Total length of bigger triangle l2 = 20 + 10 = 30 ft
l1/l2 = b1/b2
10/30 = 6/b2
b2 = 180/10
b2 = 18 ft
The ratio of the height of streetlight and man is: 18/6 = 3
Hence, the streetlight is 3 times taller than that person.

Question 17.
GEOMETRY
Use a ruler to draw two different isosceles triangles similar to the one shown. Measure the heights of each triangle.
a. Are the ratios of the corresponding heights equivalent to the ratios of the corresponding side lengths?

Answer:
b1/b2 = h1/h2
6/3 = 4/2 = 2
Hence the ratio of corresponding heights is equivalent to the ratio of corresponding side lengths.

b. Do you think this is true for all similar triangles? Explain.

Answer:
Yes, this will be true for all the similar triangles because the heights of the two similar triangles are multiplied by the same amount as the sides.

Question 18.
CRITICAL THINKING
Given △ABC ∼ △DEF and △DEF ∼ △JKL, is △ABC ∼ △JKL? Justify your answer.

Answer:
When △ABC is similar to △DEF and △DEF is similar to △JKL, then △ABC is similar to △JKL.

Lesson 2.7 Perimeters and Areas of Similar Figures

EXPLORATION 1

Work with a partner. Draw a rectangle in the coordinate plane.
a. Dilate your rectangle using each indicated scale factor k. Then complete the table for the perimeter P of each rectangle. Describe the pattern.

b. Compare the ratios of the perimeters to the ratios of the corresponding side lengths. What do you notice?
c. Repeat part(a) to complete the table for the area A of each rectangle. Describe the pattern.

d. Compare the ratios of the areas to the ratios of the corresponding side lengths. What do you notice?

e. The rectangles shown are similar. You know the perimeter and the area of the red rectangle and a pair of corresponding side lengths. How can you find the perimeter of the blue rectangle? the area of the blue rectangle?

2.7 Lesson

Try It

Question 1.
The height of Figure A is 9 feet. The height of a similar Figure B is 15 feet. What is the value of the ratio of the perimeter of A to the perimeter of B?

Answer: The ratio of the perimeter of A to B is 3/5

Explanation:
We know that when two figures are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Height of figure A/Height of figure B
Perimeter of figure A/Perimeter of figure B = 9/15 = 3/5
Thus the ratio of the perimeter of A to B is 3/5

Try It

Question 2.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters. What is the value of the ratio of the area of P to the area of Q?

Answer:
We know that when two figures are similar then the value of the ratio of their area is equal to the square of the value of the ratio of their corresponding side lengths.
The base of Triangle P is 8 meters. The base of a similar Triangle Q is 7 meters.
b1 = 8 m
b2 = 7 m
Area of triangle P/Area of triangle Q = base of triangle P/base of triangle Q

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

COMPARING PERIMETERS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the perimeters of the similar figures.

Question 3.

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
l1 = 9
l2 = 7
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 9/7
Thus the ratio of the perimeter of red to blue figure is 9/7

Question 4.

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
b1 = 8
b2 = 10
Perimeter of red figure/Perimeter of blue figure = base length of red figure/base length of blue figure
Perimeter of red figure/Perimeter of blue figure = 8/10 = 4/5
Thus the ratio of the perimeter of red to blue triangle is 4/5

COMPARING AREAS OF SIMILAR FIGURES
Find the value of the ratio (red to blue) of the areas of the similar figures.

Question 5.

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of figure B = (Side length of figure A/Side length of figure B)²
l1 = 12
l2 = 8
Perimeter of red figure/Perimeter of blue figure = side length of red figure/side length of blue figure
Perimeter of red figure/Perimeter of blue figure = (12/8)² = (3/2)² = 9/4
Thus the ratio of the perimeter of red to blue figure is 9/4

Question 6.

Answer:
We know that when two figures is similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
l1 = 12
l2 = 8
Area of red figure/Area of blue figure = side length of red figure/side length of blue figure
Area of red figure/Area of blue figure = (4/5)² = 16/225
Thus the ratio of the perimeter of red to blue triangle is 16/225

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
Two similar triangular regions are prepared for development.


a. It costs $6 per foot to install fencing. How much does it cost to surround the forest with a fence?

Answer:
Given,
It costs $6 per foot to install fencing.
Perimeter of grassland/perimeter of forest = Height of grassland/Height of forest
h1 = 60 yards
The perimeter of grassland = 240 yards
Height of forest h2 = 45 yards
Perimeter of grassland/perimeter of forest = 60/45
240/ perimeter of forest = 60/45
the perimeter of forest = 180 yards
Convert from yards to feet
180 yards = 540 feet
Thus the cost of fencing forest = 6 × 540 = $3,240

b. The cost to prepare 1 square yard of grassland is $15 and the cost to prepare 1 square yard of forest is $25. Which region costs more to prepare? Justify your answer.

Answer:
Perimeter of grassland/perimeter of forest = (Height of grassland/Height of forest)²
Height of grassland h1 = 60 yard
Height of forest h2 = 45 yards
Area of grassland = 2400 yd²
Cost to prepare 1 sq yd of grassland = $15
Cost to prepare 1 sq yd of forest = $25
Area of forest = (2400 × 9)/16
Thus the area of forest is 1350 yd²
Cost to prepare grassland = $15 × 2400 = $36,000
Cost to prepare of forest = $25 × 1350 = $33,750
Thus the grassland will cost more to prepare.

Question 8.
DIG DEEPER!
You buy a new television with a screen similar in shape to your old television screen, but with an area four times greater. The size of a television screen is often described using the distance between opposite corners of the screen. Your old television has a 30-inch screen. What is the size of your new television screen? Explain.

Answer:
Area of ΔABC/Area of ΔDEF = (Side length of AB/Side length of DE)²
Let the area of the screen of old television be x
Let the area of the screen of new television be 4x
l1 = 30 in
Area of the screen of new television/Area of the screen of new television= (distance of the screen of new television/distance of the screen of old television)²
4x/x = (distance of the screen of new television/30)²
distance of the screen of new television = 30 × 2 = 60 inch
Hence the distance of the screen of the new television is 60 inches.

Perimeters and Areas of Similar Figures Homework & Practice 2.7

Review & Refresh

The red figure is similar to the blue figure. Describe a similarity transformation between the figures.

Question 1.

Answer:
First, dilate the red figure using the scale factor of 3 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 6/2 = 3
Now reflect the image obtained after a dilation about the y-axis because both red and blue triangle is facing each other.

Question 2.

Answer:
First, dilate the red figure using the scale factor of 0.5 because the side lengths of the blue figure are 3 times the side length of the red figure.
Scale factor = 2/4 = 0.5
Then rotate the image obtained after dilation in direction 90 degrees clockwise about the origin.

Find the area of the figure.

Question 3.

Answer:
We know that the formula for the area of trapezoid = Base × height
h = 16 cm
b = 9 cm
Area of figure = 16 × 9 = 144 sq. cm
Hence the area of the given figure is 144 sq. cm

Question 4.

Answer:
h = 5 in
b = 3 in
We know that,
A = 1/2 × b × h
A = 1/2 × 5 × 3
A = 7.5 sq. cm
Thus the area of the given figure is 7.5 sq. cm

Question 5.

Answer:
h = 5 km
b1 = 6 km
b2 = 8 km
We know that,
Area of trapezoid = 1/2 × h × (b1 + b2)
A = 1/2 × 5 × 14 = 35 sq. km
Hence the area of the trapezoid is 35 sq. km

Concepts, Skills, &Problem Solving
COMPARING SIMILAR FIGURES
Dilate the figure using the indicated scale factor k. What is the value of the ratio (new to original) of the perimeters? the areas? (See Exploration 1, p. 83.)

Question 6.
a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
Given a triangle with vertices (0, 0), (0, 2), and (2, 0); k = 3
A(0, 0) = A'(0 . 3, 0 . 3) = A'(0, 0)
B(0, 2) = B'(0 . 3, 2 . 3) = B'(0, 6)
C(2, 0) = C'(2 . 3, 0 . 3) = C'(6, 0)
The coordinates of the image are A'(0, 0), B'(0, 6), C'(6, 0)
AB = √(2 – 0)² – (0 – 0)² = 2
A’B’ = √(6 – 0)² – (0 – 0)² = 6
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new triangle/Perimeter of the original triangle = Distance of A’B’/Distance of AB = 6/2 = 3
Area of new triangle/Area of the original triangle = (Distance of A’B’/Distance of AB)² = (6/2)² = 3² = 9

Question 7.
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5

Answer:
When the points of a given figure are dilated we simply multiply each x-coordinate and y-coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
where a is the scale factor
a square with vertices (0, 0), (0, 4), (4, 4), and (4, 0); k = 0.5
A(0, 0) = A'(0 . 0.5, 0 . 0.5) = A'(0, 0)
B(0, 4) = B'(0 . 0.5, 4 . 0.5) = B'(0, 2)
C(4, 4) = C'(4 . 0.5, 4 . 0.5) = C'(2, 2)
D(4, 0) = D'(4 . 0.5, 0 . 0.5) = D'(2, 0)
Coordinates of the image are A'(0, 0), B'(0, 2), C'(2, 2), D'(2, 0)
AB = √(4 – 0)² + (0 – 0)² = 4
A’B’ = √(2 – 0)² + (0 – 0)² = 2
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of new square/Perimeter of the original square = Distance of A’B’/Distance of AB = 2/4 = 1/2
Area of new square /Area of the original square = (Distance of A’B’/Distance of AB)² = (2/4)² = 1/4

PERIMETERS AND AREAS OF SIMILAR FIGURES
Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 8.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 11/6
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (11/6)² = 121/36

Question 9.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 5/8
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (5/8)² = 25/64

Question 10.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure= 4/7
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (4/7)² = 16/49

Question 11.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of red figure/Perimeter of the blue figure = Distance of red figure/Distance of blue figure = 14/9
Area of red figure /Area of the blue figure = (Distance of red figure/Distance of blue figure)² = (14/9)² = 196/81

USING SIMILAR FIGURES
The figures are similar. Find x.

Question 12.
The ratio of the perimeters is 7 : 10.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
7/10 = x/12
x = 84/10
x = 8.4
Thus the value of x is 8.4

Question 13.
The ratio of the perimeters is 8 : 5.

Answer:
We know that when two figure are similar then the value of the ratio of their perimeter is equal to the value of the ratio of their corresponding side lengths.
Perimeter of figure A/Perimeter of the figure B = Distance of figure A/Distance of figure B
8/5 = x/16
x = 25.6
Thus the value of x is 25.6

Question 14.
COMPARING AREAS
The playing surfaces of two foosball tables are similar. The ratio of the corresponding side lengths is 10:7. What is the ratio of the areas?

Answer:
Area of figure A /Area of figure B = (Distance of figure A/Distance of figure B)²
Area of figure A /Area of figure B = (10/7)²
Area of figure A /Area of figure B = 100/49
Hence, the ratio of their areas is 100/49

Question 15.
CRITICAL THINKING
The ratio of the side length of Square A to the side length of Square B is 4:9. The side length of Square A is 12 yards. What is the perimeter of Square B?

Answer:
Given,
The ratio of the side length of Square A to the side length of Square B is 4/9.
The side length of Square A is 12 yards.
side length of Square A/side length of Square B = 4/9
12 /side length of Square B = 4/9
side length of Square B = 27 yards
We know that,
The perimeter of the square is = 4s
The perimeter of the square B = 4 × 27 = 108 yards

Question 16.
MODELING REAL LIFE
The cost of the piece of fabric shown is $1.31. What would you expect to pay for a similar piece of fabric that is 18 inches by 42 inches?

Answer:
Given,
l = 21 in
w = 9 in
Area of the rectangle = l × w
A = 21 × 9 = 189 sq. in
The cost of the piece of fabric shown is $1.31
The cost of 1 sq. in of fabric = 1.31/189
l = 18 in
b = 42 in
Area of new fabric = 18 × 42 = 756 sq. in
Given the cost of new fabric = 1.31/189 × 756 = $5.24
Hence the cost of the new fabric is $5.24

Question 17.
PROBLEM SOLVING
A scale model of a merry-go-round and the actual merry-go-round are similar.

a. How many times greater is the base area of the actual merry-go-round than the base area of the scale model? Explain.

Answer:
Radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/Area of base of model merry = (120/6)² = 400

b. What is the base area of the actual merry-go-round in square feet?

Answer:
The radius of model merry go round = 6 in
Radius of actual merry go round = 10 ft = 120 in
Area of base of actual merry = 450 sq. in
Area of base of actual merry/Area of base of model merry = (Radius of actual merry/Radius of model merry)²
Area of base of actual merry/450 = (120/6)² = 400
Area of base of actual merry = 400 × 450 = 180000 sq. in = 1250 ft²

Question 18.
STRUCTURE
The circumference of Circle K is π. The circumference of Circle L is 4π. What is the value of the ratio of their circumferences? of their radii? of their areas?

Answer:
Given,
The circumference of Circle K is π
The circumference of Circle L is 4π.
circumference of Circle = 2πr
2πr = π
The radius of circle K r1 = 1/2
2πr = 4π
The radius of circle K r2 = 2
The ratio of their circumference = π/4π = 1/4
The ratio of radius of both circle = 1/4
The ratio of their area = π(r1)²/π(r2)² = 1/16

Question 19.
GEOMETRY
A triangle with an area of 10 square meters has a base of 4 meters. A similar triangle has an area of 90 square meters. What is the height of the larger triangle?

Answer:
Given,
A triangle with an area of 10 square meters has a base of 4 meters.
A similar triangle has an area of 90 square meters.
Area of the triangle = bh/2
h = 2a/b
h = (2 × 10)/4
h = 5 meters
Area of larger triangle/Area of smaller triangle = (height of larger triangle/height of smaller triangle)²
90/10 = (height of larger triangle/5)²
3 = (height of larger triangle/5)
Thus the height of larger triangle = 3 × 5 = 15 meters

Question 20.
PROBLEM SOLVING
You need two bottles of fertilizer to treat the flower garden shown. How many bottles do you need to treat a similar garden with a perimeter of 105 feet?

Answer:
The sides of the above figure are 4ft, 15 ft, 18 ft, 5 ft
Perimeter = 4ft + 15 ft + 18 ft + 5 ft = 42 ft
Number of bottle of fertilizer used in the above garden = 2
Number of bottle of fertilizer used in 1 ft = 2/42
Fertilizer used for 105 ft = 2/42 × 105 = 5
Thus 5 bottles will be used for 105 feet

Question 21.
REPEATED REASONING
Three square mirrors are used for a light reflection experiment. The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6. The ratio of the area of Mirror B to the area of Mirror C is 16 : 25. The perimeter of Mirror C is 280 centimeters. What is the area of Mirror A? Justify your answer.

Answer:
Given,
Three square mirrors are used for a light reflection experiment.
The ratio of the side length of Mirror A to the side length of Mirror B is 5 : 6 = 5/6
The ratio of the area of Mirror B to the area of Mirror C is 16 : 25 = 16/25
The perimeter of Mirror C is 280 centimeters
Side length of mirror C = Perimeter/4 = 280/4 = 70 cm
(Side length of mirror B/Side length of mirror C)² = Area of mirror A/Area of mirror B
(Side length of mirror B/Side length of mirror C)² = 16/25
(Side length of mirror B/Side length of mirror C) = 4/5
The side length of mirror B = 4/5 × Side length of mirror C
Side length of mirror B = 4/5 × 70 = 56 cm
A = s × s
A = 56 cm × 56 cm = 3136 sq. cm
Area of mirror A/Area of mirror B = (Side length of mirror A/Side length of mirror B)²
Area of mirror A/3136 = (5/6)²
Area of mirror A = 25/36 × 3136
Area of mirror A = 2177.7 sq. cm

Transformations Connecting Concepts

2 Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A scale drawing of a helipad uses a scale of 1 ft : 20 ft. The scale drawing has an area of 6.25 square feet. What is the area of the actual helipad?
Understand the problem.
You know the scale of the drawing and the area of the helipad in the drawing. You are asked to find the area of the actual helipad.
Make a plan.
A scale drawing is similar to the actual object. So, use the scale 1 ft : 20 ft and the ratio 6.25 ft² : A ft² to write and solve a proportion that represents the area A of the actual helipad.
Solve and Check.
Use the plan to solve the problem. Then check your solution.

Answer: 125 sq. ft

Question 2.
The locations of three cargo ships are shown in the coordinate plane. Each ship travels at the same speed in the same direction. After 1 hour, the x- and y-coordinates of Ship A increase 80%. Use a translation to describe the change in the locations of the ships. Then find the new coordinates of each ship.

Question 3.
All circles are similar. A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches. What is the scale factor? Justify your answer.

Answer:
Given,
A circle with a radius of 2 inches is dilated, resulting in a circle with a circumference of 22π inches.
C = 2π . r
22π = 2π . 2
p = 2π . 2
Thus the scale factor is 2.

Performance Task

Master Puppeteer

At the beginning of this chapter, you watched a STEAM Video called “Shadow Puppets.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.

Transformations Connecting Concepts

2 Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.

Graphic Organizers
You can use a Summary Triangle to explain a concept. Here is an example of Summary Triangle for translating a figure.

Choose and complete a graphic organizer to help you study the concept.

1. reflecting a figure
2. rotating a figure
3. congruent figures
4. dilating a figure
5. similar figures
6. perimeters of similar figures
7. areas of similar figures

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.

2.1 Translations (pp. 43–48)

Tell whether the blue figure is a translation of the red figure.

Question 1.

Answer: No

Explanation:
The answer is no because in the case of translation the size of the figure does not change, only the position of the figure changes. But here the size of the blue figure is larger as compared to the red figure so this is not the translation. Here the blue figure is the result of the dilation of red figure.

Question 2.

Answer: Yes

Explanation:
Yes, because in the case of translation the size of the figure does not change, only the position of the figure changes. Here the size of both blue figure and the red figure is the same but there is only a change in the position of the red figure to get blue figure. Here the blue figure is the result of the translation of the red figure.

Question 3.
The vertices of a quadrilateral are W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2). Draw the figure and its image after a translation 3 units left and 2 units down.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: W(1, 2), X(1, 4), Y(4, 4), and Z(4, 2) and a = -3, b = -2
W'(1+a, 2+b) = W'(1-3, 2-2) = W'(-2,0)
X'(1+a, 4+b) = X'(1-3, 4-2) = X'(-2,2)
Y'(4+a, 4+b) = Y'(4-3, 4-2) = Y'(1, 2)
Z'(4+a, 2+b) = C'(4-3, 2-2) = Z'(1,0)
Hence the coordinate of image are W'(-2,0), X'(-2,2), Y'(1, 2), Z'(1,0)

Question 4.
The vertices of a triangle are A(-1, -2), B(-2, 2), and C(-3, 0). Draw the figure and its image after a translation 5 units right and 1 unit up.

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(-1, -2), B(-2, 2), and C(-3, 0) and a = 5, b = 1
A'(-1+a, -2+b) = A'(-1-5, -2+1) = A'(4,-1)
B'(-2+a, 2+b) = B'(-2+5, 2+1) = B'(3,3)
C'(-3+a, 0+b) = C'(-3+5, 0+1) = C'(2, 1)
Hence the coordinate of image are A'(4,-1), B'(3,3), C'(2, 1)

Question 5.
Your locker number is 20 and your friend’s locker number is 33. Describe the location of your friend’s locker relative to the location of your locker.

Answer:
The location of my friend’s locker is first 1 locker Down and then 3 locker Right.

Question 6.
Translate the triangle 4 units left and 1 unit down. What are the coordinates of the image?

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: A(3, 5), B(6, 3), and C(4, 1) and a = -4, b = -1
A'(3+a, 5+b) = A'(3-4, 5-1) = A'(-1,4)
B'(6+a, 3+b) = B'(6-4, 3-1) = B'(2,2)
C'(4+a, 1+b) = C'(4-4, 1-1) = C'(0, 0)
Hence the coordinate of image are A'(-1,4), B'(2,2), C'(0, 0)

Question 7.
Describe a translation of the airplane from point A to point B.

Answer:
First, move the aeroplane 6 units right from point A and then 4 units down.

2.2 Reflections (pp. 49 – 54)

Tell whether the blue figure is a reflection of the red figure.

Question 8.

Answer:
No, because in the above figure the plane of reflection is inclined at 45 degrees with the horizontal line. So, the reflected figure will be perpendicular to the original figure. But in the given figure both are facing each other. This means that the blue figure is not the reflection of red figure.

Question 9.

Answer:
The answer is no because the blue figure is not the mirror image of the red figure. The side of the red figure is not facing the side of the blue figure this means that the blue figure is not the reflection of red figure.

Question 10.

Answer:
The answer is yes because the blue figure is the mirror image of the red figure. The side of the red figure is facing the side of the blue figure this means that the blue figure is the reflection of red figure.

Draw the figure and its reflection in (a) the x-axis and (b) the y-axis. Identify the coordinates of the image.

Question 11.
A(2, 0), B(1, 5), C(4, 3)

Answer:
A(x, y) = A'(x, -y)
Given: A(2, 0), B(1, 5), C(4, 3)
Reflection about x-axis:
A(2, 0) = A'(2, 0)
B(1, 5) = B'(1, -5)
C(4, 3) = C'(4, -3)

Reflection about y-axis:
A(x, y) = A'(-x, y)
A(2, 0) = A'(-2, 0)
B(1, 5) = B'(-1, 5)
C(4, 3) = C'(-4, 3)

Question 12.
D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)

Answer:
Given, D(-5, -5), E(-5, 0), F(-2, -2), G(-2, -5)
Reflection about x-axis:
A(x, y) = A'(x, -y)
D(-5, -5) = D'(-5, 5)
E(-5, 0) = E'(-5, 0)
F(-2, -2) = F'(-2, 2)
G(-2, -5) = G'(-2, 5)

Reflection about y-axis:
A(x, y) = A'(-x, y)
D(-5, -5) = D(5, -5)
E(-5, 0) = E'(5, 0)
F(-2, -2) = F'(2, -2)
G(-2, -5) = G'(2, -5)

Question 13.
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1). Find the coordinates of the figure after reflecting in the x-axis, and then translating 3 units right.

Answer:
We know that when a point is reflected about x-axis then y-coordinate becomes the opposite.
A(x, y) = A'(x, -y)
The vertices of a rectangle are E(-1, 1), F(-1, 3), G(-5, 3), and H(-5, 1).
Reflection about x-axis:
E(-1, 1) = E'(-1, -1)
F(-1, 3) = F'(-1, -3)
G(-5, 3) = G'(-5, 3)
H(-5, 1) = H'(-5, 1)
Thus the coordinates of the image are E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1)
Now translating the image 3 units Right.
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if shift is Right and Vertical Up and the value of a and b will be negative if shift is left and vertical Down.
Given: E'(-1, -1), F'(-1, -3), G'(-5, 3), H'(-5, 1) a = 3, b = 0
E”(-1 + a, -1 + b) = E”(-1 + 3, -1 + 0) = E”(2, -1)
F”(-1 + a, -3 + b) = F”(-1 + 3, -3 + 0) = F”(2, -3)
G”(-5 + a, 3 + b) = G”(-5 + 3, 3 + 0) = G”(-2, 3)
H”(-5 + a, 1 + b) = H”(-5 + 3, 1 + 0) = H”(-2, 1)
Thus the coordinates of the image are E”(2, -1), F”(2, -3), G”(-2, 3), H”(-2, 1)

The coordinates of a point and its image after a reflection are given. Identify the line of reflection.

Question 14.
(-1, -3) → (1, -3)

Answer:
Given,
(-1, -3) → (1, -3)
We can see that the y-coordinate of both points and its image are the same but the x-coordinate of the image is the opposite of its points.
Hence, Y-axis is the line of reflection.

Question 15.
(2, 1) → (2, -1)

Answer:
Given,
(2, 1) → (2, -1)
We can see that the x-coordinate of both points and its image are the same but the y-coordinate of the image is the opposite of its points.
Hence, X-axis is the line of reflection.

Question 16.
You perform an experiment involving angles of refraction with a laser pen. You point a laser pen from point L at a mirror along the red path and the image is a reflection in the y-axis.
a. Does the light reach a cat at point C? Explain.

Answer:
Yes, the light will reach at point C.
Because the coordinate of point L is (4, 3) and the coordinate of point C is (-4, 3) and it is given problems that laser is reflected about the y-axis.
So when point L(4, 3) is reflected about the y-axis its x-coordinates become opposite and y-coordinates remain the same.
So when point L(4, 3) is reflected about the y-axis its image will be point C(-4, -3)

b. You bounce the light off the mirror so its path is a reflection. What line of reflection is needed for the light to reach the cat?

Answer: The line of reflection will be y-axis.

2.3 Rotations (pp. 55–62)

Tell whether the blue figure is a rotation of the red figure about the origin. If so, give the angle and the direction of rotation.

Question 17.

Answer:
The answer is no because the blue figure is the mirror image of the red figure. The blue figure is the result of the reflection of red figure about the y-axis. Also, both red and blue figure are facing each other with the y-axis in the center of both which remains that it is not the case of rotation.

Question 18.

Answer:
The answer is yes, because the red figure is in the 1st quadrant and the blue figure in third quadrant. Also, both blue and red figure are facing each other in opposite directions which blue figure is the result of the rotation of red figure.
When red figure is rotated 180 degrees counterclockwise it will result in a blue figure.

The vertices of a triangle are A(-4, 2), B(-2, 2), and (-3, 4). Rotate the triangle as described. Find the coordinates of the image.

Question 19.
180° about the origin

Answer:
We know that when a point is rotated 180 degrees about origin then both x coordinate and y coordinate becomes opposite.
A(x, y) = A'(-x, -y)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotated 180 degrees about origin:
A(-4, 2) = A'(4, -2)
B(-2, 2) = B'(2, -2)
C(-3, 4) = C'(3, -4)
The coordinate of the image are A'(4, -2), B'(2, -2), C'(3, -4)

Question 20.
270° clockwise about the origin

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Given points: A(-4, 2), B(-2, 2), and C(-3, 4)
Rotating 270 degrees clockwise about the origin:
A(-4, 2) = A'(-2, -4)
B(-2, 2) = B'(-2, -2)
C(-3, 4) = C'(-4, -3)
The coordinate of the image are A'(-2, -4), B'(-2, -2), C'(-4, -3)

Question 21.
A bicycle wheel is represented in a coordinate plane with the center of the wheel at the origin. Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6). After a bike ride, the reflectors have rotated 90° counterclockwise about the origin. What are the locations of the reflectors at the end of the bike ride?

Answer:
We know that when a point is rotated 90 degrees counterclockwise about origin then both x coordinate and y coordinate becomes opposite.
P(x, y) = P'(-y, x)
Reflectors are placed on the bicycle wheel at points (7, 4) and (-5, -6)
A(7, 4) = A'(-4, 7)
B(-5, -6) = B'(6, -5)
Hence the new coordinate of the reflector are A'(-4, 7), B'(6, -5)

2.4 Congruent Figures (pp. 63–68)

Identify any congruent figures in the coordinate plane.

Question 22.

Answer:
AB = ED
BC = DC
CA = CE
∠A = ∠E
∠B = ∠D
∠C = ∠C
When we see both triangles ABC and EDC closely we observe that both the triangles are the mirror images of each other with the y-axis as the line of reflection. So all the corresponding sides are equal and also all the corresponding angles, this means that both the triangle are congruent.
Hence ΔABC is congruent to ΔEDC
GH = JK
HF = KI
FG = IJ
∠G = ∠J
∠H = ∠K
∠F = ∠I
So we can see that all the corresponding sides are equal and also all the corresponding angles. this means that both the triangles are congruent.
Hence ΔGHF is congruent to ΔJKI

Question 23.

Answer:
When we observe square ABCD and square EFGH we can see that
AB = EF
BC = FG
CD = GH
DA = HE
∠A = ∠E
∠B = ∠F
∠C = ∠G
∠D = ∠H
We can see that all the corresponding sides are equal and also all the corresponding angles, this means that both are congruent.
RS = IJ
ST = JK
TU = KL
UR = LI
∠R = ∠I
∠S = ∠J
∠T = ∠K
∠L = ∠U
Hence rectangle RSTU is congruent to rectangle IJKL

The red figure is congruent to the blue figure. Describe a sequence of rigid motions between the figures.

Question 24.

Answer:
First, rotate the blue figure 90° clockwise because the blue figure is in a vertical position but the red figure is in the horizontal position.
Translate the image 5 units right because the first image formed after the rotation will in the second quadrant but the red figure in the first quadrant.

Question 25.

Answer:
First, reflect the blue figure about the y-axis because both red and blue figure is facing each other and they are the mirror image of each other.
Then translate the image 2 units up because the first image formed after reflection will be at the same distance from the x-axis but the red figure is touchung the x-axis.

Question 26.
The figures are congruent. Name the corresponding angles and the corresponding sides.

Answer:
Corresponding sides are
AB = KL
BC = LM
CA = MK
Corresponding angles
∠A = ∠K
∠B = ∠L
∠C = ∠M

Question 27.
Trapezoids EFGH and QRST are congruent.
a. What is the length of side QR ?

Answer:
Length of side:
QR = EF = 3 feet

b. Which angle in QRST corresponds to ∠H?
Answer:
The angle that corresponds to ∠H is ∠T

c. What is the perimeter of QRST ?

Answer:
Perimeter of QSRT = QR + RS + ST + TQ
= EF + FG + GH + HE
= 3 + 5 + 4 + 8
= 20 ft

2.5 Dilations (pp. 69–76)

Tell whether the blue figure is a dilation of the red figure.

Question 28.

Answer:
The answer is no because dilation the size of the image either increases or decreases that depend on the type of dilation. We can see that both the red and blue figure are of the same size which means that the blue figure is not dilation of the red figure.

Question 29.

Answer:
The answer is yes because both red and blue figures are of the same shape and all the corresponding angles are equal but the blue figure is larger than the red figure. And in dilation, the size of the image is either increases or decreases but the shape always remains the same. So blue figure is the result of dilation of red figure.

The vertices of a figure are given. Draw the figure and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 30.
P(-3, -2), Q(-3, 0), R(0, 0); k = 4

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
P(-3, -2), Q(-3, 0), R(0, 0); k = 4
P(-3, -2) = P'(-3 . 4, -2 . 4) = P'(-12, -8)
Q(-3, 0) = Q'(-3 . 4, 0 . 4) = Q'(-12, 0)
R(0, 0) = R'(0 . 4, 0 . 4) = R'(0, 0)
Thus the coordinates of the image: P'(-12, -8), Q'(-12, 0), R'(0, 0)

Question 31.
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)

Answer:
A(x, y) = A'(x . a, y . a)
Where a is the scaling factor
Given,
B(3, 3), C(3, 6), D(6, 6), E(6, 3); k = \(\frac{1}{3}\)
B(3, 3) = B'(3 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = B'(1, 1)
C(3, 6) = C'(3 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = C'(1, 2)
D(6, 6) = D'(6 . \(\frac{1}{3}\), 6 . \(\frac{1}{3}\)) = D'(2, 2)
E(6, 3) = E'(6 . \(\frac{1}{3}\), 3 . \(\frac{1}{3}\)) = E'(2, 1)
Thus the coordinates of the image: B'(1, 1), C'(1, 2), D'(2, 2), E'(2, 1)

Question 32.
The blue figure is a dilation of the red figure. Identify the type of dilation and find the scale factor.

Answer:
AB = 1 unit
A’B’ = 2 units
scale factor = size of image figure/size of actual figure
scale factor = 2/1 = 2
We can see from the above figure that the size of the image figure are larger as compared to the size of the original image so it is the Enlargement dilation.

Question 33.
The vertices of a rectangle are Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4). Dilate the rectangle with respect to the origin using a scale factor of \(\frac{3}{2}\). Then translate it 5 units right and 1 unit down. What are the coordinates of the image?

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the rectangle: Q(-6, 2), R(6, 2), S(6, -4), and T(-6, -4), scale factor = \(\frac{3}{2}\)
Q(-6, 2) = Q'(-6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = Q'(-9, 3)
R(6, 2) = R'(6 . \(\frac{3}{2}\), 2 . \(\frac{3}{2}\)) = R'(9, 3)
S(6, -4) = S'(6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = S'(9, 6)
T(-6, -4) = T'(-6 . \(\frac{3}{2}\), -4 . \(\frac{3}{2}\)) = T'(-9, -6)
Thus the coordinates of the image: Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6)
Q'(-9, 3), R'(9, 3), S'(9, 6), T'(-9, -6) a = 5, b = -1
Q”(-9 + a, 3 + b) = Q”(-9 + 5, 3 – 1) = Q”(-4, 2)
R”(9 + a, 3 + b) = R”(9 + 5, 3 – 1) = R”(14, 2)
S”(9 + a, -6 + b) = S”(9 + 5, -6 – 1) = S”(14, -7)
T”(-9 + a, -6 + b) = T”(-9 + 5, -6 – 1) = T”(-4, -7)
Thus the coordinates of the image: Q”(-4, 2), R”(14, 2), S”(14, -7), T”(-4, -7)

2.6 Similar Figures

Question 34.
Determine whether the two figures are similar. Explain your reasoning.

Answer:
No, the above two figures are not similar.

Question 35.
Draw figures with the given vertices in a coordinate plane. Which figures are similar? Explain your reasoning.
Triangle A: (-4, 4), (-2, 4), (-2, 0)
Triangle B: (-2, 2), (-1, 2), (-1, 0)
Triangle C: (6, 6), (3, 6), (3, 0)

Answer:
Triangle A: (-4, 4), (-2, 4), (-2, 0)

Triangle B: (-2, 2), (-1, 2), (-1, 0)

Triangle C: (6, 6), (3, 6), (3, 0)

The figures are similar. Find x.

Question 36.

Answer:
Ratio of sides of larger triangle = Ratio of sides of smaller triangle
20/14 = x/7
x = 10
Thus the value of x is 7 inches.

Question 37.

Answer:
Ratio of sides of larger parallelogram= Ratio of sides of smaller parallelogram
x/6 = 6/4
x = 9
Thus the value of x is 9 cm

2.7 Perimeters and Areas of Similar Figures (pp. 83-88)

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 38.

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 6/8 = 3/4
Hence the ratio of perimeter of red triangle to blue is 3/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (6/8)² = 9/16
Thus the ratio of the perimeter of the red to the blue triangle is 9/16

Question 39.

Answer:
Perimeter of figure A/Perimeter of figure B = Side length of figure A/Side length of figure B
Perimeter of red figure/Perimeter of blue figure = 28/16 = 7/4
Hence the ratio of the perimeter of red rectangle to blue is 7/4
Area of figure A/Area of figure B = (Side length of A/Side length of B)²
Area of red figure/Area of blue figure = (28/16)² = 49/16
Thus the ratio of the perimeter of the red to the blue rectangle is 49/16

The figures are similar. Find x.

Question 40.
The ratio of the perimeters is 5 :7.

Answer:
5/7 = 12/x
x =(12 × 7)/5
x = 16.8 cm
Thus the value of x is 16.8 cm

Question 41.
The ratio of the perimeters is 6 : 5.

Answer:
6/5 = x/6
x = 36/5
x = 7.2
Thus the value of x is 7.2

Question 42.
Two photos are similar. The ratio of the corresponding side lengths is 3 : 4. What is the ratio of the areas?

Answer:
Area of photo A/Area of photo B = (Side length of photo A/Side length of photo B)²
Area of photo A/Area of photo B = (3/4)² = 9/16
Thus the ratio of the area of two photos is 9/16

Question 43.
The ratio of side lengths of Square A to Square B is 2 : 3. The perimeter of Square A is 16 inches. What is the area of Square B?

Answer:
Given,
The ratio of side lengths of Square A to Square B is 2 : 3.
The perimeter of Square A is 16 inches.
Perimeter of Square A/Perimeter of Square B = Side length of Square A/Side length of Square B
The perimeter of Square B = (16 × 3)/2
The perimeter of Square B = 24 inches
Side length of square B = Perimeter/4 = 24/4 = 6 inch
Area of square B = s × s = 6 × 6 = 36 sq. in

Question 44.
The TV screen is similar to the computer screen. What is the area of the TV screen?

Answer:
Area of computer screen = 108 sq. in
The side length of the computer screen = 12 in
The side length of the TV screen = 20 in
Area of TV screen = (25 × 108)/9
Area of TV screen = 300 sq. in
Hence the area of the TV screen is 300 sq. in

Transformations Practice Test

2 Practice Test

Triangles ABC and DEF are congruent.

Question 1.
Which angle of DEF corresponds to ∠C ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Thus the angle correspond to ∠C is ∠F

Question 2.
What is the perimeter of DEF ?

Answer:
Corresponding sides
Side AB = Side DE
Side BC = Side EF
Side CA = Side ED
Corresponding angles
∠A = ∠D
∠B = ∠E
∠C = ∠F
Perimeter of DEF = DE + EF + FD
= AB + BC + CA
= 5 + 4 + 6
= 15 cm
Thus the perimeter of the ΔDEF is 15 cm

Tell whether the blue figure is a translation, reflection, rotation, or dilation of the red figure.

Question 3.

Answer: The scale factor of a dilation is greater than 1 because the shape of the blue figure is larger than the red figure.

Question 4.

Answer:
The blue figure is the reflection of the red figure. Because the blue figure is the mirror image of red figure. Also, the shape and size of both red and blue figures are the same.

Question 5.

Answer:
The blue figure is the translation of the red figure. Because the shape and size of both red and blue figures are the same. And both the figure is not facing to each other. This means that the blue figure is the result of the translation of the red figure.

Question 6.

Answer:
The blue figure is the result of the rotation of the red figure. Because the shape and size of both the red and blue figure are the same but the figure is horizontal and the blue figure is vertical. This means that the blue figure is the result of the rotation of the term figure.

The vertices of a triangle are A(2, 5), B(1, 2), and C(3, 1). Find the coordinates of the image after the transformations given.

Question 7.
Reflect in the y-axis.

Answer:
A(x, y) = A'(-x, y)
A(2, 5), B(1, 2), C(3, 1)
Reflection about the y-axis:
A(2, 5) = A'(-2, 5)
B(1, 2) = B'(-1, 2)
C(3, 1) = C'(-3, 1)
Thus the coordinates of the image: A'(-2, 5), B'(-1, 2), C'(-3, 1)

Question 8.
Rotate 90° clockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)

Question 9.
Reflect in the x-axis, and then rotate 90° counterclockwise about the origin.

Answer:
A(x, y) = A'(x, -y)
Given, A(2, 5), B(1, 2), C(3, 1)
A(2, 5) = A'(2, -5)
B(1, 2) = B'(1, -2)
C(3, 1) = C'(3, -1)
Thus the coordinates of the image: A'(2, -5), B'(1, -2), C'(3, -1)
A'(2, -5) = A”(2, 5)
B'(1, -2) = B”(1, 2)
C'(3, -1) = C”(3, 1)
Thus the coordinates of the image: A”(2, 5), B”(1, 2), C”(3, 1)

Question 10.
Dilate with respect to the origin using a scale factor of 2. Then translate 2 units left and 1 unit up.

Answer:
When the points of given figure are dilated we simply multiply each x and y coordinate by the given scale factor.
P(x, y) = P'(x . a, y . a)
Given points of the triangle: A(2, 5), B(1, 2), C(3, 1) and scale factor = 2
Dilating the figure by scale factor of 2
A (2, 5) = A'(2 . 2, 5 . 2) = A'(4, 10)
B (1, 2) = B'(1 . 2, 2 . 2) = B'(2, 4)
C (3, 1) = C'(3 . 2, 1 . 2) =  C'(6, 2)
Hence the coordinates of the image are A'(4, 10), B'(2, 4),  C'(6, 2)
Now translating image 2 unit left and 1 unit up.
Given: A'(4, 10), B'(2, 4),  C'(6, 2) a = -2, b = 1
A”(4 + a, 10 + b) = A”(4 – 2, 10 + 1) = A”(2, 11)
B”(2 + a, 4 + b) = B”(2 – 2, 4 + 1) = B”(0, 5)
C”(6 + a, 2 + b) = C”(6 – 2, 2 + 1) = C”(4, 3)
Hence the coordinates of the image are A”(2, 11), B”(0, 5), C”(4, 3)

Question 11.
In a coordinate plane, draw Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2); Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1); and Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3). Which figures are similar? Explain your reasoning.

Answer:
Rectangle A: (-4, 4), (0, 4), (0, 2), (-4, 2)

Rectangle B: (-2, 2), (0, 2), (0, 1), (-2, 1)

Rectangle C:(-6, 6), (0, 6), (0, 3), (-6, 3)

Question 12.
Translate a point (x, y) 3 units left and 5 units up. Then translate the image 5 units right and 2 units up. What are the coordinates of the point after the translations?

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
Given: A(x,y) and a = -3, b = 5
A'(x+a, y+b) = A'(x – 3, y + 5)
Now translating image 5 units right and 2 units up.
Image after first translation: A'(x – 3, y + 5) and a = 5, b = 2
A”(x – 3 + a, y + 5 + b) = A”(x – 3 + 5, y + 5 + 2) = A”(x + 2, y + 7)
Thus the final image will be A”(x + 2, y + 7)

Question 13.
The two figures are similar.
(a) Find the value of x.

Answer:
Ratio of sides of red figure = Ratio of sides of blue figure
x/14 = 10/8
x = (10 × 14)/8
x = 17.5

(b) Find the values of the ratios (red to blue) of the perimeters and of the areas.

Answer:
Perimeter of red figure/Perimeter of blue figure = Side length of red figure/Side length of blue figure
Perimeter of red figure/Perimeter of blue figure = 14/8 = 7/4
Thus the ratio of the perimeter of red to blue figure is 7/4
Area of red figure/Area of blue figure = (side length of red figure/side length of blue figure)²
Area of red figure/Area of blue figure = (14/8)² = 49/16
Thus the ratio of the perimeter of red to blue triangle is 49/16

Question 14.
A wide-screen television measures 36 inches by 54 inches. A movie theater screen measures 42 feet by 63 feet. Are the screens similar? Explain.

Answer:
Given,
A wide-screen television measures 36/54 = 2/3
A movie theater screen measures 42/63 = 2/3
We can see that the ratio of corresponding sides of the television screen is equal to the ratio of corresponding sides of the movie theatre. So television screens and movie theatres are similar.

Question 15.
You want to use the rectangular piece of fabric shown to make a pair of curtains for your window. Name the types of congruent shapes you can make with one straight cut. Draw an example of each type.

Answer:
The types of congruent shapes that can be made with one straight cut
2 right triangles
2 rectangles
2 right trapezoid

Transformations Cumulative Practice

Cumulative Practice

Question 1.
A clockwise rotation of 90° is equivalent to a counterclockwise rotation of how many degrees?

Answer:
90° of clockwise rotation = (360 – 90)° of counterclockwise rotation
= 270° of counterclockwise rotation

Question 2.
The formula K = C + 273.15 converts temperatures from degrees Celsius C to Kelvin K. Which of the following formulas is not correct?
A. K – C = 273.
B. C = K – 273.15
C. C – K = -273.15
D. C = K + 273.15

Answer: C = K + 273.15

Question 3.
You want to solve the equation -3(x + 2) = 12x. What should you do first?
F. Subtract 2 from each side.
G. Add 3 to each side.
H. Multiply each side by -3.
I. Divide each side by -3.

Answer: I. Divide each side by -3.

Explanation:
-3(x + 2) = 12x
x + 2 = -4x
x = -4x – 2
x + 4x = -2
5x = -2
x = –\(\frac{2}{5}\)
Thus the correct answer is option I.

Question 4.
Which value of x makes the equation \(\frac{3}{4} x\) = 12 true?
A. 9
B. 11\(\frac{1}{4}\)
C. 16
D. 48

Answer: C. 16

Explanation:
\(\frac{3}{4} x\) = 12
3x = 12 × 4
3x = 48
x = \(\frac{48}{3}\)
x = 16
Thus the correct answer is option C.

Question 5.
A triangle is graphed in the coordinate plane. What are the coordinates of the image after a translation 3 units right and 2 units down?

F. A'(1, 4), B'(1, 1), C'(3, 1)
G. A'(1, 2), B'(1, -1), C'(3, -1)
H. A'(-2, 2), B'(-2, -1), C'(0, -1)
I. A'(0, 1), B'(0, -2), C'(2, -2)

Answer:
We know that to translate a figure ‘a’ units horizontally and ‘b’ units vertically in the coordinate plane, ‘a’ is added to x-coordinate and ‘b’ is added to the y-coordinate of the vertices.
A(x,y) = A'(x+a, y+b)
The value a and b will be positive if the shift is Right and Vertical Up and the value of a and b will be negative if the shift is left and vertical Down.
A(-2, 4), B(-2, 1), C(0, 1) and a = 3, b = -2
A'(-2+a, 4+b) = A'(-2 + 3, 4 – 2) = A'(1, 2)
B'(-2+a, 1+b) = B'(-2 + 3, 1 – 2) = B'(1,-1)
C'(0+a, 1+b) = C'(0 + 3, 1 – 2) = C'(3, -1)
Coordinate of the image are: A'(1, 2), B'(1,-1), C'(3, -1)
Thus the correct answer is option G.

Question 6.
Your friend solved the equation in the box shown. What should your friend do to correct the error that he made?

Answer:
–\(\frac{x}{3}\) + \(\frac{2}{5}\) = –\(\frac{7}{15}\)
–\(\frac{x}{3}\) = –\(\frac{13}{15}\)
x = 2\(\frac{3}{5}\)
Thus the correct answer is option C.

Question 7.
Your teacher dilates the rectangle using a scale factor of \(\frac{1}{2}\).

What is the area of the dilated rectangle in square inches?

Answer:
l = 10 in
b = 6 in
scale factor = \(\frac{1}{2}\)
New length after dilation = 10 × \(\frac{1}{2}\) = 5
New breadth after dilation = 6 × \(\frac{1}{2}\) = 3
Area of rectangle = l × b
A = 5 × 3 = 15 sq. in
The area of the dilated rectangle will be 5 in²

Question 8.
Your cousin earns $9.25 an hour at work. Last week she earned $222.00 How many hours did she work last week?
F. \(\frac{1}{24}\)
G. 22 hours
H. 24 hours
I. 212.75 hours

Answer: H. 24 hours

Explanation:
Given,
Your cousin earns $9.25 an hour at work.
Last week she earned $222.00
Total no. of working hour = total earning of week/rate of one hour
= \(\frac{222}{9.25}\)
= 24 hours
Thus the correct answer is option H.

Question 9.
Triangle EFG is a dilation of Triangle HIJ. Which proportion is not true for Triangle EFG and Triangle HIJ ?

Answer: \(\frac{EG}{HI}\) = \(\frac{FG}{IJ}\)
The correct answer is option B.

Question 10.
The red figure is congruent to the blue figure. Which of the following is a sequence of rigid motions between the figures?
F. Reflect the red triangle in the x-axis, and then translate it 3 units left.
G. Reflect the red triangle in the x-axis, and then translate it 3 units right.
H. Reflect the red triangle in the y-axis, and then translate it 3 units left.
I. Rotate the red triangle 90° clockwise about the origin.

Answer:
1. First red triangle is reflected about the x-axis because both red and blue triangles are the mirror image of each other and also the red triangle is in the 1st quadrant and the blue triangle is in 4th quadrant.
2. Then translate the image 3 unit left because the base of both red and blue triangles is not opposite to each other.
Thus the correct answer is option F.

Question 11.
Several transformations are used to create the pattern.
Part A
Describe the transformation of Triangle GLM to Triangle DGH

Answer:
Both ΔGLM and Δ DGH are of the same shape and size but their position are different so the transformation will be translated.

Part B
Describe the transformation of Triangle ALQ to Triangle GLM.

Answer:
The size of the triangle ALQ is four times the size of triangle GLM and the shape of both triangles is the same so the transformation will be dilation.

Part C
Triangle DFN is a dilation of Triangle GHM. Find the scale factor.

Answer:
The size of the triangle DFN is double the size of triangle GHM. So the scale factor of dilation will be 2.

Question 12.
A rectangle is graphed in the coordinate plane.

What are the coordinates of the image after a reflection in the y-axis?
A. J'(4, 1), K'(4, 3), L'( 1, 3), M'(-1, 1)
B. J'(-4, 1), K(-4, -3), L'(1, -3), M'(1, 1)
C. J'(1, 4), K'(3, 4), L'(3, -1), M'(1, -1)
D. J'(-4, 1), K'(-4, 3), L'(1, 3), M'(1, 1)

Answer:
We know that when a point is reflected about the y-axis then the x-coordinate becomes the opposite.
A(x, y) = A'(-x, y)
J(-4, 1), K(-4, 3), L(1, 3), M(1, 1)
Reflection about the y-axis:
J(-4, 1) = J'(4, 1)
K(-4, 3) = K'(4, 3)
L(1, 3) = L'(-1, 3)
M(1, 1) = M'(-1, 1)
Coordinate of image are: J'(4, 1), K'(4, 3), L'(-1, 3), M'(-1, 1)
Thus the correct answer is option A.

Conclusion:

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Big Ideas Math Book 8th Grade Answer Key Chapter 4 Graphing and Writing Linear Equations

Make the most out of the handy resources available for Big Ideas Math Ch 4 Graphing and Writing Linear Equations and stand out from the rest of the crowd. BIM Book 8th Grade Chapter 4 Solutions include questions belonging to Lessons 4.1 to 4.7, Cumulative Practice, Assessment Tests, Review Tests, etc. Big Ideas Math 8th Grade Chapter 4 Solution Key is given by subject experts after extensive research. Access the quick links over here during your preparation and get the assistance needed at the comfort of your home.

Performance Task

• Graphing and Writing Linear Equations STEAM Video/Performance Task
• Graphing and Writing Linear Equations Getting Ready for Chapter 4

Lesson: 1 Graphing Linear Equations

• Lesson 4.1 Graphing Linear Equations
• Graphing Linear Equations Homework & Practice 4.1

Lesson: 2 Slope of a Line

• Lesson 4.2 Slope of a Line
• Slope of a Line Homework & Practice 4.2

Lesson: 3 Graphing Proportional Relationships

• Lesson 4.3 Graphing Proportional Relationships
• Graphing Proportional Relationships Homework & Practice 4.3

Lesson: 4 Graphing Linear Equations in Slope-Intercept Form

• Lesson 4.4 Graphing Linear Equations in Slope-Intercept Form
• Graphing Linear Equations in Slope-Intercept Form Homework & Practice 4.4

Lesson: 5 Graphing Linear Equations in Standard Form

• Lesson 4.5 Graphing Linear Equations in Standard Form
• Graphing Linear Equations in Standard Form Homework & Practice 4.5

Lesson: 6 Writing Equations in Slope-Intercept Form

• Lesson 4.6 Writing Equations in Slope-Intercept Form
• Writing Equations in Slope-Intercept Form Homework & Practice 4.6

Lesson: 7 Writing Equations in Point-Slope Form

• Lesson 4.7 Writing Equations in Point-Slope Form
• Writing Equations in Point-Slope Form Homework & Practice 4.7

Chapter: 4 – Graphing and Writing Linear Equations

• Graphing and Writing Linear Equations Connecting Concepts
• Graphing and Writing Linear Equations Chapter Review
• Graphing and Writing Linear Equations Practice Test
• Graphing and Writing Linear Equations Cumulative Practice

Graphing and Writing Linear Equations STEAM Video/Performance Task

STEAM Video

“Hurricane
A hurricane is a storm with violent winds. How can you prepare your home for a hurricane?
Watch the STEAM Video “Hurricane!” Then answer the following questions.

1. Robert says that the closer you are to the eye of a hurricane, the stronger the winds become. The wind speed on an island is 50 miles per hour when the eye of a hurricane is 140 miles away.
a. Describe the wind speed on the island when the eye of the hurricane is 100 miles away.
b. Describe the distance of the island from the eye of the hurricane when the wind speed on the island is 25 miles per hour.
c. Sketch a line that could represent the wind speed y (in miles per hour) on the island when the eye of x the hurricane is miles away from the island. Wind speed
2. A storm dissipates as it travels over land. What does this mean?

Performance Task

Anatomy of a Hurricane
After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about the atmospheric pressure inside a hurricane.

You will be asked to use a model to find the strength of a hurricane after x hours of monitoring. Why is it helpful to predict how strong the winds of a hurricane will become?

Graphing and Writing Linear Equations Getting Ready for Chapter 4

Chapter Exploration
1. Work with a partner.
a. Use the equation y = \(\frac{1}{2}\)x + 1 to complete the table. (Choose any two x-values and find the y-values.)
b. Write the two ordered pairs given by the table. These are called solutions of the equation.

c. PRECISION Plot the two solutions. Draw a line exactly through the points.
d. Find a different point on the line. Check that this point is a solution of the equation y = \(\frac{1}{2}\)x + 1.
e. LOGIC Do you think it is true that any point on the line is a solution of the equation y = \(\frac{1}{2}\)x + 1? Explain.
f. Choose five additional x-values for the table below. (Choose both positive and negative x-values.) Plot the five corresponding solutions. Does each point lie on the line?

g. LOGIC Do you think it is true that any solution of the equation y = \(\frac{1}{2}\)x + 1 is a point on the line? Explain.
h. Why do you think y = ax + b is called a linear equation?

Vocabulary
The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
linear equation
slope
y-intercept
solution of a linear equation
x-intercept

Lesson 4.1 Graphing Linear Equations

EXPLORATION 1

Creating Graphs
Work with a partner. It starts snowing at midnight in Town A and Town B. The snow falls at a rate of 1.5 inche sper hour.
a. In Town A, there is no snow on the ground at midnight. How deep is the snow at each hour between midnight and 6 A.M.? Make a graph that represents this situation.

b. Repeat part(a) for TownB, which has 4 inches of snow on the ground at midnight.
c. The equations below represent the depth y(in inches) of snow x hours after midnight in Town C and Town D. Graph each equation.
Town C y = 2x + 3
Town D y = 8

d. Use your graphs to compare the snowfall in each town.
Answer:

Try It

Graph the linear equation.
Question 1.
y = 3x
Answer:
Make to table of values
Replace x with a number and find the value of y

Plot the values of x and y obtained above, on the graph

Draw the line through the points

Question 2.
y = – 2x – 1
Answer:

Plot the values of x and y

Now the line through the points

Question 3.
y = –\(\frac{1}{2}\)x + 2
Answer:

Plot the ordered pairs

Graph the linear equation.
Question 4.
y = 3
Answer:
The graph of y = 3 is a horizontal like passing through (0, 3)
Draw a horizontal line through this point.

Question 5.
y = – 1.5
Answer:
The graph of y = -1.5 is a horizontal line passing through (0, -1.5)
Draw a horizontal line through this point.

Question 6.
x = – 4
Answer:
The graph of x = – 4 is a vertical line passing through (-4, 0)
Draw a vertical line through this point.

Question 7.
x = \(\frac{1}{2}\)
Answer:
The graph of x = \(\frac{1}{2}\) is a vertical line passing through (\(\frac{1}{2}\), 0)
Draw a vertical line through this point.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

GRAPHING A LINEAR EQUATION Graph the linear equation.
Question 8.
y = – x + 1
Answer:
Make of a table of values
Replace x with a number and find the value of y

Plot the values of x and y obtained, on the graph,

Question 9.
y = 0.8x – 2
Answer:
Replace x with a number and find the value of y

Question 10.
x = 2.5
Answer:
The graph of x = 2.5 is a vertical line passing through (2.5, 0)
Draw a vertical line through this point.

Question 11.
y = \(\frac{2}{3}\)
Answer:
The graph of y = \(\frac{2}{3}\) is a horizontal line passing through (0, \(\frac{2}{3}\))
Draw a horizontal line through this point.

Question 12.
WHICH ONE DOESN’T BELONG?
Which equation does not belong with the other three? Explain your reasoning.

Answer:
y = x – 2
4x + 3 = y
y = x² + 6
x + 5 = y

Self-Assessment for Problem Solving
Solve each exercise. The rate your understanding of the success criteria in your journal.

Question 13.
A game show contestant earns y dollars for completing a puzzle in x minutes. This situation is represented by the equation y = – 250x + 5000. How long did a contestant who earned $500 take to complete the puzzle? Justify your answer.
Answer:
Given,
A game show contestant earns y dollars for completing a puzzle in x minutes.
This situation is represented by the equation y = – 250x + 5000.
y = -250x + 5000
500 = -250x + 5000
500 – 5000 = -250x + 5000 – 5000
-4500 = -250x
x = 18

Question 14.
The total cost y (in dollars) to join a cheerleading team and attend x competitions is represented by the equation y = 10x + 50.

a. Graph the linear equation.

Answer:

b. You have $75 to spend. How many competitions can you attend?
Answer:
75 ≤ 10x + 50
75 – 50 ≤ 10x
25 ≤ 10x
2.5 ≥ x
By this I can say that I can attend 2 competitions if I have $75 to spend.

Question 15.
The seating capacity for a banquet hall is represented by y = 8x + 56, where x is the number of extra tables you need. How many extra tables do you need to double the original seating capacity?
Answer:
Given,
The seating capacity for a banquet hall is represented by y = 8x + 56, where x is the number of extra tables you need.
y = 8x + 56
2 × 56 = 8x + 56
112 = 8x + 56
8x = 112 – 56
8x = 56
x = 7 tables

Graphing Linear Equations Homework & Practice 4.1

Review & Refresh

Tell whether the triangles are similar. Explain.
Question 1.

Answer:
x° + 46° + 95° = 180°
x° + 141° = 180°
x° = 180° – 141°
x° = 39°
Thus the angles of the triangle are 39°, 46°, 95°
y° + 39° + 46° = 180°
y° + 75° = 180°
y° = 180° – 75°
y° = 95°
Three angles of the triangle are 39°, 46°, 95°
The triangles have two pairs of congruent angles.

Question 2.

Answer:
x° + 40° + 51° = 180°
x° + 91° = 180°
x° = 180° – 91°
x° = 89°
Three angles of the triangle are 40°, 51°, 89°
y° + 40° + 79° = 180°
y° + 119° = 180°
y° = 180° – 119°
y° = 61°

Describe the translation of the point to its image.
Question 3.
(1, – 4) → (3, 0)
Answer:
A(1, -4) = A'(1 + 2, -4) = (3, -4)
A'(3, 4) = B(3, -4 + 4) = (3, 0)
Translate 2 units right and 4 units up.

Question 4.
(6, 4) → (- 4, – 6)
Answer:
We are given the points
(6, 4) → (- 4, – 6)
A(6, 4) = A'(6 – 10, 4) = (-4, 4)
A'(-4, -4) = B(-4, 4 – 10) = (-4, -6)

Question 5.
(4, – 2) → (- 9, 3)
Answer:
We are given the points
A(4, -2)
B(-9, 3)
A(4, -2) = A'(4 – 13, -2) = (-9, -2)
A'(-9, -2) = B(-9, -2 + 4) = (-9, 3)

Concepts, Skills, & Problem Solving

CREATING GRAPHS Make a graph of the situation. (See Exploration 1, p. 141.)
Question 6.
The equation y = – 2x + 8 represents the amount (in fluid ounces) of dish detergent in a bottle after x days of use.
Answer:

Question 7.
The equation y = 15x + 20 represents the cost (in dollars) of a gym membership after x months.
Answer:

PRECISION Copy and complete the table with two solutions. Plot the ordered pairs and draw the graph of the linear equation. Use the graph to find a third solution of the equation.
Question 8.

Answer:

(x, y) = (2, 5)

Question 9.

Answer:

(x, y) = (3, 3)

GRAPHING A LINEAR EQUATION Graph the linear equation.
Question 10.
y = – 5x
Answer:

Question 11.
y = 9x
Answer:

Question 12.
y = 5
Answer:
The graph of y = 5 is a horizontal line passing through (0, 5)
Draw a horizontal line through this point.

Question 13.
x = – 6
Answer:

Question 14.
y = x – 3
Answer:

Question 15.
y = – 7x – 1
Answer:

Question 16.
y = – \(\frac{x}{8}\) + 4
Answer:

Question 17.
y = 0.75x – 0.5
Answer:

Question 18.
y = – \(\frac{2}{3}\)
Answer:

Question 19.
y = 6.75
Answer:

Question 20.
x = – 0.5
Answer:
The graph of x = -0.5 is a vertical line passing through (-0.5, 0)
Draw a vertical line through this point.

Question 21.
x = \(\frac{1}{4}\)
Answer:
The graph of x = \(\frac{1}{4}\) is a vertical line passing through (\(\frac{1}{4}\), 0)
Draw a vertical line through this point.

Question 22.
YOU BE THE TEACHER
Your friend graphs the equation y = 4. Is your friend correct? Explain your reasoning.
Answer:

No my friend is not correct because the graph for the equation y = 4 is a  horizontal line not a vertical line, and it passes through the point (0, 4) not (4, 0)

Question 23.
MODELING REAL LIFE
The equation y = 20 represents the cost y (in dollars) for sending x text messages in a month. Graph the linear equation. What does the graph tell you about your texting plan?

Answer:

Question 24.
MODELING REAL LIFE
The equation y = 2x + 3 represents the cost y (in dollars) of mailing a package that weighs x pounds.

a. Use a graph to estimate how much it costs to mail the package.
b. Use the equation to find exactly how much it costs to mail the package.
Answer:
Given the equation y = 2x + 3
The ordered pairs will be (0, 3), (2,7), (4, 11)
Now plot the ordered pairs

y = 2(1.126) + 3
= 5.252 ≈ 5.25

SOLVING A LINEAR EQUATION Solve for y. Then graph the linear equation.
Question 25.
y – 3x = 1
Answer:
y – 3x = 1
y = 3x + 1

Draw a line through the points

Question 26.
5x + 2y = 4
Answer:
5x + 2y = 4
2y = 4 – 5x
y = – \(\frac{5}{2}\)x + 2

Question 27.
– \(\frac{1}{3}\)y + 4x = 3
Answer:
– \(\frac{1}{3}\)y + 4x = 3
– \(\frac{1}{3}\)y = 3 – 4x
y = 12x – 9

Question 28.
x + 0.5y = 1.5
Answer:
x + 0.5y = 1.5
0.5y = -x + 1.5
y = -2x + 3

Question 29.
MODELING REAL LIFE
The depth y (in inches) of a lake after x years is represented by the equation y = 0.2x + 42. How much does the depth of the lake increase in four years? Use a graph to justify your answer.

Answer:
y = 0.2x + 42
Depth of the lake now: y = 0.2(0) + 42 = 42
Depth of the lake after 4 years: y = 0.2(4) + 42 = 42.8

42.8 – 42 = 0.8 inches

Question 30.
MODELING REAL LIFE
The amount y (in dollars) of money in your savings account after x months is represented by the equation y = 12.5x + 100.
a. Graph the linear equation.

Answer:

b. How many months will it take you to save a total of $237.50?
Answer:
y = 12.5x + 100
237.5 = x + 100
237.5 – 100 = 12.5x + 100 – 100
12.5x = 137.5
x = 11

Question 31.
PROBLEM SOLVING
The radius y (in millimeters) of a chemical spill after x days is represented by the equation y = 6x + 50.

a. Graph the linear equation.

Answer:

b. The leak is noticed after two weeks. What is the area of the leak when it is noticed? Justify your answer.
Answer:
y = 6(14) + 50
y = 84 + 50
y = 134 mm
2πr = 2π = 841.95 sq. mm

Question 32.
GEOMETRY
The sum S of the interior angle measures of a polygon with n sides is S = (n – 2) • 180°.
a. Plot four points (n, S) that satisfy the equation. Is the equation a linear equation? Explain your reasoning.

Answer:

b. Does the value n = 3.5 make sense in the context of the problem? Explain your reasoning.
Answer:
The value n = 3.5 does not make sense because the number of angles cannot be other than integer greater or equal to 2.

Question 33.
DIG DEEPER!
One second of video on your cell phone uses the same amount of memory as two pictures. Your cell phone can store 2500 pictures.
a. Create a graph that represents the number y of pictures your cell phone can store when you take x seconds of video.

Answer:

b. How many pictures can your cell phone store in addition to a video that is one minute and thirty seconds long?
Answer:
Determine the number of pictures you can store in addition to a video of 1 min 30 seconds.
1 min 30 seconds = (60 + 90) 3 seconds = 90 seconds
2500 – (2 . 90)
2500 – 180 = 2320 pictures

Lesson 4.2 Slope of a Line

EXPLORATION 1

Measuring the Steepness of a Line
Work with a partner. Draw any nonvertical line in a coordinate plane.
a. Develop a way to measure the steepness of the line. Compare your method with other pairs.
b. Draw a line that is parallel to your line. What can you determine about the steepness of each line? Explain your reasoning.
Answer:

EXPLORATION 2

Using Right Triangles
Work with a partner. Use the figure shown.

a. △ABC is a right triangle formed by drawing a horizontal line segment from point A and a vertical line segment from point B. Use this method to draw another right triangle, △DEF, with its longest side on the line
b. What can you conclude about the two triangles in part(a)? Justify your conclusion. Compare your results with other pairs.
c. Based on your conclusions in part(b), what is true about \(\frac{BC}{AC}\) and the corresponding measure in △DEF? Explain your reasoning. What do these values tell you about the line?
Answer:

Try It

Find the slope of the line.
Question 1.

Answer:
(x1, y1) = (-2, 3)
(x2, y2) = (3, 2)
m = (y2 – y1)/(x2 – x1)
m = (2 -3)/(3 – (-2))
m = -1/5
Thus slope = -1/5

Question 2.

Answer:
(x1, y1) = (-4, -1)
(x2, y2) = (2, 1)
m = (y2 – y1)/(x2 – x1)
m = (1 – (-1))/(2 – (-4))
m = 2/6
Thus slope = 1/3

Find the slope of the line through the given points.
Question 3.
(1, -2), (7, -2)
Answer:
(x1, y1) = (1, -2)
(x2, y2) = (7, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – (-2))/(7 – 1)
m = 0/6
Thus slope = 0

Question 4.
(-3, -3), (-3, -5)
Answer:
(x1, y1) = (-3, -3)
(x2, y2) = (-3, -5)
m = (y2 – y1)/(x2 – x1)
m = (-5 + 3)/(-3 + 3)
m = -2/0
Thus slope = undefined

Question 5.
WHAT IF
The blue line passes through (-4, -3) and (-3, 2). Are any of the lines parallel? Explain.
Answer:
(x1, y1) = (-4, -3)
(x2, y2) = (-3, 2)
m = (y2 – y1)/(x2 – x1)
m = (2 + 3)/(-3 + 4)
m = 5/1
m = 5
The slpe of the blue line is 5 and the slope of the red line is also 5.
The blue lines and red lines have same slopes so they are parallel.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
VOCABULARY
What does it mean for a line to have a slope of 4?
Answer:
If a line have a slope of 4 it means that the line rises 4 units for every 1 units it runs.

FINDING THE SLOPE OF A LINE Find the slope of the line through the given points.
Question 7.
(1, -1), (6, 2)
Answer:
(x1, y1) = (1, -1)
(x2, y2) = (6, 2)
m = (y2 – y1)/(x2 – x1)
m = (2 – (-1))/(6 – 1)
m = 3/5

Question 8.
(2, -3), (5, -3)
Answer:
(x1, y1) = (2, -3)
(x2, y2) = (5, -3)
m = (y2 – y1)/(x2 – x1)
m = (5 – 2)/(-3 + 3)
m = 3/0
m = undefined

Question 9.
FINDING SLOPE
Are the lines parallel? Explain your reasoning.

Answer:
Red line:
(x1, y1) = (-1, 0)
(x2, y2) = (1, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 0)/1 – (-1))
m = -2/2
m = -1
Blue Line:
(x1, y1) = (-1, 3)
(x2, y2) = (1, -1)
m = (y2 – y1)/(x2 – x1)
m = (-1 – 3)/(1 – (-1))
m = -4/2
m = -2
The slope of the blue line and red line are not the same. So they are not parallel.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
The table shows the lengths (in inches) of your hair months after your last haircut. The points in the table lie on a line. Find and interpret the slope of the line. After how many months is your hair 4 inches long?

Answer:
Determine the slope of the line using two points from the table:
(2, 1), (4, 2)
m = (2 – 1)/4 – 2
m = 1/2
m = 0.5
This means that each month the hair grows 0.5 inches
As the hair grows 0.5 inches/ month, it will be 4 inches long after 4/0.5 = 8 months.

Question 11.
A customer pays an initial fee and a daily fee to rent a snowmobile. The total payment for 3 days is 92 dollars. The total payment for 5 days is 120 dollars. What is the daily fee? Justify your answer.
Answer:
Given,
A customer pays an initial fee and a daily fee to rent a snowmobile.
The total payment for 3 days is 92 dollars. The total payment for 5 days is 120 dollars.
m = (120 – 92)/5 – 3
m = 28/2
m = 14

Question 12.
You in-line skate from an elevation of 720 feet to an elevation of 750 feet in 30 minutes. Your friend in-line skates from an elevation of 600 feet to an elevation of 690 feet in one hour. Compare your rates of change in elevation.

Answer:
Given,
You in-line skate from an elevation of 720 feet to an elevation of 750 feet in 30 minutes.
Your friend in-line skates from an elevation of 600 feet to an elevation of 690 feet in one hour.
(750 – 720)/30 = 30/30 = 1 ft/min
(690 – 600)/60 = 90/60 = 1.5 ft/min

Slope of a Line Homework & Practice 4.2

Review & Refresh

Graph the linear equation.
Question 1.
y = 4x – 3
Answer:

Question 2.
x = -3
Answer:

Question 3.
y = 2
Answer:

Question 4.
y = \(\frac{3}{2}\)x – \(\frac{1}{2}\)
Answer:

Find the missing values in the ratio table.
Question 5.

Answer:

x/10 = 1/3
x = 10/3
x = 3.33
1/3 = 5/y
y = 5 × 3
y = 15
1/3 = 7/z
z = 3 × 7
z = 21

Question 6.

Answer:

Concepts, Skills, &Problem Solving

USING RIGHT TRIANGLES Use the figure shown (See Exploration 2, p. 147.)

Question 7.
Find the slope of the line.
Answer:
(x1, y1) = B(-4, 2)
(x2, y2) = A(-2, 1)
m = (y2 – y1)/(x2 – x1)
m = (1 – 2)/(-2 – (-4))
m = -1/2
Thus the slope m = -1/2

Question 8.
Let point D be at (-4, 1). Use the sides of △BDA to find the slope of the line.
Answer:

m = -BD/DA = -1/2

FINDING THE SLOPE OF A LINE Find the slope of the line.
Question 9.

Answer:
(x1, y1) = (-2, 0)
(x2, y2) = (2, 3)
m = (y2 – y1)/(x2 – x1)
m = (3 – 0)/(2 – (-2))
m = 3/4

Question 10.

Answer:
(x1, y1) = (-2, 5)
(x2, y2) = (2, 0)
m = (y2 – y1)/(x2 – x1)
m = (0 – 5)/(2 – (-2))
m = -5/4

Question 11.

Answer:
(x1, y1) = (-4, 1)
(x2, y2) = (1, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 1)/(1 + 2)
m = -3/5

Question 12.

Answer:
(x1, y1) = (-5, -4)
(x2, y2) = (1, -3)
m = (y2 – y1)/(x2 – x1)
m = (-3 – (-4))/(1 – (-5))
m = 1/6

Question 13.

Answer:
(x1, y1) = (-1, 3)
(x2, y2) = (3, 3)
m = (y2 – y1)/(x2 – x1)
m = (3 – 3)/(3 – (-1))
m = 0/4
m = 0

Question 14.

Answer:
(x1, y1) = (1, 3)
(x2, y2) = (1, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 3)/(1 – 1)
m = -5/0
m = undefined

FINDING THE SLOPE OF A LINE Find the slope of the line through the given points.
Question 15.
(4, -1), (-2, -1)
Answer:
(x1, y1) = (4, -1)
(x2, y2) = (-2, -1)
m = (y2 – y1)/(x2 – x1)
m = (-1 – (-1))/(-2 – 4)
m = 0/-6
m = 0

Question 16.
(5, -3), (5, 8)
Answer:
(x1, y1) = (5, -3)
(x2, y2) = (5, 8)
m = (y2 – y1)/(x2 – x1)
m = (8 – 3)/(5 – 5)
m = 5/0
m = undefined

Question 17.
(-7, 0), (-7, -6)
Answer:
(x1, y1) = (-7, 0)
(x2, y2) = (-7, -6)
m = (y2 – y1)/(x2 – x1)
m = (-6 – 0)/(-7 – (-7))
m = -6/0
m = undefined

Question 18.
(-3, 1), (-1, 5)
Answer:
(x1, y1) = (-3, 1)
(x2, y2) = (-1, 5)
m = (y2 – y1)/(x2 – x1)
m = (5 – 1)/(-1 + 3)
m = 4/2
m = 2

Question 19.
(10, 4), (4, 15)
Answer:
(x1, y1) = (10, 4)
(x2, y2) = (4, 15)
m = (y2 – y1)/(x2 – x1)
m = (15 – 4)/(4 – 10)
m = 11/-6
m = -11/6

Question 20.
(-3, 6), (2, 6)
Answer:
(x1, y1) = (-3, 6)
(x2, y2) = (2, 6)
m = (y2 – y1)/(x2 – x1)
m = (6 – 6)/(2 – (-3))
m = 0/5
m = 0

Question 21.
REASONING
Draw a line through each point using slope of m = \(\frac{1}{4}\). Do the lines intersect? Explain.

Answer:

The 2 lines are parallel because they have the same slope and they do not intersect.

Question 22.
YOU BE THE TEACHER
Your friend finds the slope of the line shown. Is your friend correct? Explain your reasoning.

Answer:

No my friend is not correct because the denominator should be 2 – 4
(x1, y1) = (2, 3)
(x2, y2) = (4, 1)
m = (y2 – y1)/(x2 – x1)
m = (1 – 3)/(4 – 2)
m = -2/2
m = -1

IDENTIFYING PARALLEL LINES Which lines are parallel? How do you know?
Question 23.

Answer:
Blue line:
(x1, y1) = (-5, 2)
(x2, y2) = (-4, -1)
m = (y2 – y1)/(x2 – x1)
m = (-1 – 2)/(-4 – (-5))
m = -3/1
m = -3
Red line:
(x1, y1) = (-2, 1)
(x2, y2) = (-1, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 1)/(-1 – (-2))
m = -3/1
m = -3
Green Line:
(x1, y1) = (1, 3)
(x2, y2) = (2, -1)
m = (y2 – y1)/(x2 – x1)
m = (-1 – 3)/(2 – 1)
m = -4/1
m = -4
Blue line and red line have slope of -3, so they are parallel.

Question 24.

Answer:
Blue line:
(x1, y1) = (-2, 3)
(x2, y2) = (-5, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 3)/(-5 – (-2))
m = -5/-3
m = 5/3
Red line:
(x1, y1) = (1, 2)
(x2, y2) = (-2, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 2)/(-2 – 1)
m = -4/-3
m = 4/3
Green Line:
(x1, y1) = (4, 1)
(x2, y2) = (1, -3)
m = (y2 – y1)/(x2 – x1)
m = (-3 – 1)/(1 – 4)
m = -4/-3
m = 4/3
Red line and green line have slope of 4/3 by this we can say that they are parallel.

IDENTIFYING PARALLEL LINES Are the given lines parallel? Explain your reasoning.
Question 25.
y = -5, y = 3
Answer:

Both lines are horizontal and have slope = 0

Question 26.
y = 0, x = 0
Answer:

The line y = 0 have slope = 0 and are horizontal lines.
The line x = 0 have slope = undefined and are vertical lines.
So, they are not parallel.

Question 27.
x = -4, x = 1
Answer:

Both lines are vertical and have an undefined slope.

FINDING SLOPE The points in the table lie on a line. Find the slope of the line.
Question 28.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (10 – 2)/(3 – 1) = (18 – 10)/(5 – 3) = (26 – 18)/(7 – 5)
m = 8/2 = 8/2 = 8/2
m = 4 = 4 = 4
Slope = 4

Question 9.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (2 – 0)/(2 – (-3)) = (4 – 2)/(7 – 2) = (6 – 4)/(12 – 7)
m = 2/5 = 2/5 = 2/5
m = 2/5

Question 30.
MODELING REAL LIFE
Carpenters refer to the slope of a roof as the pitch of the roof. Find the pitch of the roof.

Answer:
Pitch of the roof = rise/run
= 4/12 = 1/3

Question 31.
PROJECT
The guidelines for a wheelchair ramp suggest that the ratio of the rise to the run be no greater than 1 : 12.

a. CHOOSE TOOLS Find a wheelchair ramp in your school or neighborhood. Measure its slope. Does the ramp follow the guidelines?

Answer:
rise/run < 1/12
m = 0.06
1/12 = 0.0833
0.06 < 0.0833
As m < 1/12 the wheelchair ramp follows the guides.

b. Design a wheelchair ramp that provides access to a building with a front door that is 2.5 feet above the sidewalk. Illustrate your design.
Answer:
AC/AB = 1/12
2.5/AB = 1/12
AB = 2.5 × 12
AB = 30
So the end of the ramp should be placed at least 30 feet from the front door.

USING AN EQUATION Use an equation to find the value of k so that the line that passes through the given points has the given slope.
Question 32.
(1, 3), (5, k); m = 2
Answer:
A(1, 3)
B(5, k)
m = 2
2 = (k – 3)/(5 – 1)
2 × 4 = k – 3
8 = k – 3
k = 8 + 3
k = 11

Question 33.
(-2, k), (2, 0); m = -1
Answer:
Given,
A(-2, k)
B(2, 0)
m = -1
-1 = (0 – k)/2 – (-2)
-1 = -k/4
-4 = -k
k = 4

Question 34.
(-4, k), (6, -7); m = –\(\frac{1}{5}\)
Answer:
Given,
A(-4, k)
B(6, -7)
m = –\(\frac{1}{5}\)
–\(\frac{1}{5}\) = (-7 – k)/6 – (-4)
-2 = -7 – k
-2 + 7 = -k
5 = -k
k = -5

Question 35.
(4, -4), (k, -1); m = \(\frac{3}{4}\)
Answer:
\(\frac{3}{4}\) = (-1 – (-4))/(k – 4)
4 = k – 4
k = 4 + 4
k = 8

Question 36.
MODELING REAL LIFE
The graph shows the numbers of prescriptions filled over time by a pharmacy.

a. Find the slope of the line.
Answer:
(0, 0), (20, 5)
m = (5 – 0)/(20 – 0)
m = 5/20
m = 1/4
b. Explain the meaning of the slope as a rate of change.
Answer:
This means that every 4 minutes a prescription is filled.

Question 37.
CRITICAL THINKING
Which is steeper: the boatramp, or a road with a 12% grade? Note: Explain. (Road grade is the vertical increase divided by the horizontal distance.)

Answer:
Mramp = rise/run = 6/36 = 1/6
Mroad = 12% = 12/100 = 0.12
0.16 = 0.166… > 0.12
Mramp > Mroad
Therefore the slope of the ramp is steeper than the slope of the road.

Question 38.
REASONING
Do the points A(-2, -1), B(1, 5), and C(4, 11) lie on the same line? Without using a graph, how do you know?
Answer:
Given,
A(-2, -1), B(1, 5), and C(4, 11)
mAB = (5 – (-1))/(1 – (-2)) = 6/3 = 2
mBC = (11 – 5)/(4 – 1) = 6/3 = 2
By seeing the slopes we can say that the points A, B, C lie on the same line.

Question 39.
PROBLEM SOLVING
A small business earns a profit of $6500 in January and $17,500 in May. What is the rate of change in profit for this time period? Justify your answer.
Answer:
Pjan = 6500
Pmay = 17,500
Pmay – Pjan/5 – 1
= (17,500 – 6500)/4
= 11,000/4 = 2750

Question 40.
STRUCTURE
Choose two points in the coordinate plane. Use the slope formula to find the slope of the line that passes through the two points. Then find the slope using the formula \(\frac{y_{1}-y_{2}}{x_{1}-x_{2}}\). Compare your results.
Answer:
P1(2, 5)
P2(3, 10)
m1 = (10 – 5)/(3 – 2) = 5/1 = 5
m2 = (5 – 10)/(1 – 3) = -5/-1 = 5
m1 = m2

Question 41.
DIG DEEPER!
The top and the bottom of the slide are level with the ground, which has a slope of 0.

a. What is the slope of the main portion of the slide?
b. Describe the change in the slope when the bottom of the slide is only 12 inches above the ground. Explain your reasoning.
Answer:
18 inches = 1.5 feet
mMC = rise/run = (8 – 1.5)/(12 – 1 – 1) = 6.5/10 = 0.65
AD = 1
mMC = CR/MR
= (8 – 1)/(12 – 1 – 1) = 7/10 = 0.7
The slope increases from 0.65 to 0.70 because the rise increasses, while the run stays the same.

Lesson 4.3 Graphing Proportional Relationships

EXPLORATION 1

Using a Ratio Table to Find Slope
Work with a partner. The graph shows amounts of vinegar and water that can be used to make a cleaning product.
a. Use the graph to make a ratio table relating the quantities. Explain how the slope of the line is represented in the table.

b. Make a ratio table that represents a different ratio of vinegar to water. Use the table to describe the slope of the graph of the new relationship.
Answer:

EXPLORATION 2

Deriving an Equation
Work with a partner. Let (x, y) represent any point on the graph of a proportional relationship.

a. Describe the relationship between the corresponding side lengths of the triangles shown in the graph. Explain your reasoning.
b. Use the relationship in part(a) to write an equation relating y, m, and x. Then solve the equation for y. How can you find the side lengths of the triangles in the graph?
c. What does your equation in part(b) describe? What does represent? Explain your reasoning.
Answer:

Try It

Question 1.
WHAT IF
The cost of frozen yogurt is represented by y = 0.75x. Graph the equation and interpret the slope.
Answer:
The equation shows that the slope m is 0.75. So the graph passes through the points (0, 0) and (1, 0.75).
Plot the ordered pairs and draw the graph.

The slope indicates that the unit cost is $0.75 per ounce.

Question 2.
How much would a spacecraft that weighs 3500 kilograms on Earth weigh on Titan?
Answer:
y = 1/7 x
y = 1/7 × 3500
y = 500 kg
So a spacecraft would weigh 500 kg on Titan.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

GRAPHING A PROPORTIONAL RELATIONSHIP Graph the equation.
Question 3.
y = 4x
Answer:

Question 4.
y = -3x
Answer:

Question 5.
y = 8x
Answer:

Question 6.
WRITING AND USING AN EQUATION
The number of objects a x machine produces is proportional to the time (in minutes) that the machine runs. The machine produces five objects in four minutes.
a. Write an equation that represents the situation.

Answer:
As 5 objects are produced in 4 minutes, the slope of the line is m = 5/4.
The equation that represents the situation is
y = 5/4 x
y = 1.25 x

b. Graph the equation in part (a) and interpret the slope.

Answer:
Use the slope. The equation shows that the slope m is 1.25. So the graph passes through the points (0, 0) and (1, 1.25)

c. How many objects does the machine produce in one hour?
Answer:

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
The amount y (in liters) of water that flows over a natural waterfall in x seconds is represented by the equation y = 500x. The graph shows the number of liters of water that flow over an artificial waterfall. Which waterfall has a greater flow? Justify your answer.

Answer:
Given the equation y = 500x
15000 – 3000 = 12000
12000/4 = 3000
Mnatural = 500
3000 > 500
Therefore the artificial waterfall has greater flow.

Question 8.
The speed of sound in air is 343 meters per second. You see lightning and hear thunder 12 seconds later.
a. Is there a proportional relationship between the amount of time that passes and your distance from a lightning strike? Explain.

Answer:
y = kx
where k is the speed of sound, x the time and y the distance.
Yes, there is a proportional relationship between the amount of time that passes and your distance from the lightning strike as the further you are, the more time will pass until the sound reaches you.

b. Estimate your distance from the lightning strike.
Answer:
y = 343 × 12
= 4116 meters

Graphing Proportional Relationships Homework & Practice 4.3

Review & Refresh

Find the slope of the line.
Question 1.

Answer:
(x1, y1) = (0, -3)
(x2, y2) = (3, 0)
m = (y2 – y1)/(x2 – x1)
m = (0 – (-3))/(3 – 0)
m = (0 + 3)/(3 – 0)
m = 3/3
m = 1

Question 2.

Answer:
(x1, y1) = (0, 1)
(x2, y2) = (3, -5)
m = (y2 – y1)/(x2 – x1)
m = (-5 – 1)/(3 – 0)
m = -6/3
m = -2

Question 3.

Answer:
(x1, y1) = (0, 0)
(x2, y2) = (2, 8)
m = (y2 – y1)/(x2 – x1)
m = (8 – 0)/(2 – 0)
m = 8/2
m = 4

Solve the equation. Check your solution.
Question 4.
2x + 3x = 10
Answer:
Given the equation
2x + 3x = 10
5x = 10
x = 10/5
x = 2

Question 5.
x + \(\frac{1}{6}\) = 4 – 2x
Answer:
Given the equation
x + \(\frac{1}{6}\) = 4 – 2x
x + 2x = 4 – \(\frac{1}{6}\)
3x = 4 – \(\frac{1}{6}\)
3x = \(\frac{23}{6}\)
x = \(\frac{23}{18}\)

Question 6.
2(1 – x) = 11
Answer:
2(1 – x) = 11
2 – 2x = 11
2 – 11 = 2x
2x = -9
x = -9/2

Concepts, Skills, & Problem Solving

USING EQUIVALENT RATIOS The graph shows amounts of water and flour that can be used to make dough. (See Exploration 1, p. 155.)

Question 7.
Use the graph to make a ratio table relating the quantities. Explain how the slope of the line is represented in the table.
Answer:

m = rise/run
= (10 – 5)/(6 – 3)
= 5/3
That means to every 5 cups of flour there is an increase of 3 cups of water.
The slope m is 5/3.

Question 8.
Make a ratio table that represents a different ratio of flour to water. Use the table to describe the slope of the graph of the new relationship.
Answer:

From the table we find that for every increase of 7 cups of flour there is an increase of 4 cups of water.
The slope is 7/4.

Question 9.
GRAPHING AN EQUATION
The amount y(in dollars) that you raise by selling fundraiser tickets is represented by the equation y = 5x. Graph the equation and interpret the slope.
Answer:

The slope indicates that the unit cost is $5 per ticket.

IDENTIFYING PROPORTIONAL RELATIONSHIPS Tell whether and are in a proportional relationship. Explain your reasoning. If so, write an equation that represents the relationship.
Question 10.

Answer:
The graph doesn’t represent a proportional relationship because it doesn’t pass through the point (0, 0).

Question 11.

Answer:
The graph represents a proportional relationship because it is linear and passes through the point (0, 0)
(0, 0), (2, 8)
m = (8 – 0)/(2 – 0)
m = 8/2
m = 4
The equation is y = 4x

Question 12.

Answer:
(2 – 1)/(6 – 3) = 1/3
(3 – 2)/(9 – 6) = 1/3
(4 – 3)/(12 – 9) = 1/3
As the rate of change is constant, it means that the graph is a line.
(1 – y)/(3 – 0) = 1/3
(1 – y)/3 = 1/3
1 – y = 1
y = 1 – 1
y = 0
Therefore the point (0, 0) belomgs to the graph.
So the table represents a proportional relationship
y = 1/3 x

Question 13.

Answer:
(8 – 4)/(5 – 2) = 4/3
(13 – 8)/(8 – 5) = 5/3
(23 – 13)/10 – 8 = 10/2 = 5

Question 14.
MODELING REAL LIFE
The cost y (in dollars) to rent a kayak is proportional to the number x of hours that you rent the kayak. It costs $27 to rent the kayak for 3 hours.

a. Write an equation that represents the situation.
b. Interpret the slope of the graph of the equation.
c. How much does it cost to rent the kayak for 5 hours? Justify your answer.
Answer:
y = kx
27 = k × 3
k = 27/3
k = 9
The equation is k = 9x
b. The slope k = 3 shows that the cost of renting the kayak per hour is $9.
c. y = 9 × 5
y = 45

Question 15.
MODELING REAL LIFE
The distance y (in miles) that a truck travels on x gallons of gasoline is represented by the equation y = 18x. The graph shows the distance that a car travels.

a. Which vehicle gets better gas mileage? Explain how you found your answer.

Answer:
y = 18x
(0, 0), (2, 50)
m = (50 – 0)/(2 – 0)
m = 50/2
m = 25
25 > 18
Therefore the car has better mileage.

b. How much farther can the vehicle you chose in part(a) travel on 8 gallons of gasoline?
Answer:
y = 25 × 8 – 18 × 8
= 200 – 144
= 56 miles

Question 16.
PROBLEM SOLVING
Toenails grow about 13 millimeters per year. The table shows fingernail growth.

a. Do fingernails or toenails grow faster? Explain.

Answer:
y = 0.25x
m = (1.4 – 0.7)/(2 – 1)
m = 0.7
y = 0.7x
Because 0.7 > 0.25, the fingernails grow faster.

b. In the same coordinate plane, graph equations that represent the growth rates of toenails and fingernails. Compare and interpret the steepness of each graph.
Answer:

Question 17.
REASONING
The quantities and are in a proportional relationship. What do you know about the ratio of y to x for any point (x, y) on the graph of x and y?
Answer:
y = kx
where k is constant
y/x = k
This means the ratio of y to x is constant.

Question 18.
DIG DEEPER!
The graph relates the temperature change y (in degrees Fahrenheit) to the altitude change x (in thousands of feet).

a. Is the relationship proportional? Explain.

Answer: The relationship is proportional because the graph is linear and passes through the origin.

b. Write an equation of the line. Interpret the slope.

Answer:
(0,0), (10, -35)
m = (-35 – 0)/(10 – 0)
= -35/10
= -3.5
y = -3.5x

c. You are at the bottom of a mountain where the temperature is 74°F. The top of the mountain is 5500 feet above you. What is the temperature at the top of the mountain? Justify your answer.
Answer:
x = 5.5 – 0 = 5.5 thousand feet
y = -3.5x = -3.5(5.5) = -19.25
74 – 19.25 = 54.75°F

Question 19.
CRITICAL THINKING
Consider the distance equation d = rt, where d is the distance (in feet), r is the rate (in feet per second), and t is the time (in seconds). You run for 50 seconds. Are the distance you run and the rate you run at proportional? Use a graph to justify your answer.
Answer:
d = rt
d = 50r
Having the form y = kx the equation represents a proportional relationship.

Lesson 4.4 Graphing Linear Equations in Slope-Intercept Form

EXPLORATION 1

Deriving an Equation
Work with a partner. In the previous section, you learned that the graph of a proportional relationship can be represented by the equation y = mx, where m is the constant of proportionality.

a. You translate the graph of a proportional relationship 3 units up as shown below. Let (x, y) represent any point on the graph. Make a conjecture about the equation of the line. Explain your reasoning.

b. Describe the relationship between the corresponding side lengths of the triangles. Explain your reasoning.
c. Use the relationship in part(b) to write an equation relating y, m, and x. Does your equation support your conjecture in part(a)? Explain.
d. You translate the graph of a proportional relationship b units up. Write an equation relating y, m, x, and b. Justify your answer.
Answer:

Try It

Find the slope and the y-intercept of the graph of the linear equation.
Question 1.
y = 3x – 7
Answer:
Given the equation
y = 3x – 7
Write the equation in slope – intercept form: y = mx + b
The slope of the line is m and the y – intercept of the line is b.
y = 3x – 7
Slope = 3 and y – intercept = -7

Question 2.
y – 1 = –\(\frac{2}{3}\)x
Answer:
Write the equation in slope – intercept form: y = mx + b
The slope of the line is m and the y – intercept of the line is b.
y – 1 = –\(\frac{2}{3}\)x
y = –\(\frac{2}{3}\)x + 1
Slope = –\(\frac{2}{3}\) and y – intercept = 1

Graph the linear equation. Identify the x-intercept.
Question 3.
y = x – 4
Answer:
y = x – 4
Comparing the above equation with slope – intercept equation.
slope = 1, y-intercept = -4
Ploy y – intercept and slope
slope = rise/run = 1/1
Plot the point that is 1 unit right and 1 unit up from (0, -4) = (1, -3)

Thus the intercept is 4.

Question 4.
y = –\(\frac{1}{2}\)x + 1
Answer:
y = –\(\frac{1}{2}\)x + 1
Comparing the above equation with slope – intercept equation.
Slope = –\(\frac{1}{2}\), y-intercept = 1
y-intercept = 1. So plot (0, 1)
Slope = rise/run = -1/2
Plot the point that is 2 units right and 1 unit down from (0, -4) = (2, 0)

So, the x-intercept is 2.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
IN YOUR OWN WORDS
Consider the graph of the equation y = mx + b.
a. How does changing the value of m affect the graph of the equation?

Answer:
The value of m is the slope of the graph. If the value of m changes it means the slope of the graph is changing, whether it will rise or fall from left or right is dependent on the value of m.

b. How does changing the value of b affect the graph of the equation?
Answer:
The value of b is the y-intercept of the graph. If the value of b changes it means it affects where the graph crosses the y – axis.

IDENTIFYING SLOPE AND y-INTERCEPT Find the slope and the y-intercept of the graph of the linear equation.
Question 6.
y = -x + 0.25
Answer:
y = mx + c
slope = -1 and y – intercept = 0.25

Question 7.
y – 2 = –\(\frac{3}{4}\)x
Answer:
Given the equation
y – 2 = –\(\frac{3}{4}\)x
y = –\(\frac{3}{4}\)x + 2
slope = –\(\frac{3}{4}\) and y – intercept = 2

GRAPHING A LINEAR EQUATION Graph the linear equation. Identify the x-intercept.
Question 8.
y = x – 7
Answer:

The line crosses the x-axis at (7, 0)
So, the x – intercept is 7.

Question 9.
y = 2x + 8
Answer:

The line crosses the x – axis at (-4, 0)
So, the x – intercept is -4.

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
The height y (in feet) of a movable bridge after rising for seconds is represented by the equation y = 3x + 6. Graph the equation. Interpret the y-intercept and slope. How many seconds does it take the bridge to reach a height of 76 feet? Justify your answer.

Answer:
Given,
y = 3x + 6.
slope = 3, y – intercept = 16

The y – intercept is 16. So, the initial height of the bridge is 16 feet.
The slope is 3. So, the bridge rises 3 feet per second.
The bridge will reach a height of 76 feet in 20 seconds.

Question 11.
The number of perfume bottles in storage after x months is represented by the equation y = -20x + 460. Graph the equation. Interpret the y-intercept and the slope. In how many months will there be no perfume bottles left in storage? Justify your answer.

Answer:
Given the equation
y = -20x + 460
Slope = -20, y-intercept = 460

The y-intercept is 460. So, the initial number of perfume in the storage is 460.
The slope is -20. So, the number of perfume bottle decrease with 20 bottles per months.
There will be no perfume bottle left in the storage in 23 months.

Graphing Linear Equations in Slope-Intercept Form Homework & Practice 4.4

Review & Refresh

Tell whether x and y are in a proportional relationship. Explain your reasoning. If so, write an equation that represents the relationship.
Question 1.

Answer:
(8 – 6)/(2 – 1) = 2/1 = 2
(10 – 8)/(3 – 2) = 2/1 = 2
(12 – 10)/(4 – 3) = 2/1 = 2
The rate of change in the table is constant.
(6 – y)/(1 – 0) = 2
6 – y = 2
y = 6 – 2
y = 4
Therefore the graph does not pass through the origin.
So x and y are not proportional.

Question 2.

Answer:
(4 – 0)/(-8 – 0) = 4/-8 = -1/2 = -0.5
(2 – 4)/(-4 – (-8)) = -2/4 = -1/2 = -0.5
(-2 – 2)/(4 – (-4)) = -4/8 = -1/2 = -0.5
(-4 – (-2))/(8 – 4) = -2/4 = -1/2 = -0.5
As the rate of change is constant, x and y are in a proportional relationship.
y = -0.5x

Solve the equation for y.
Question 3.
x = 4y – 2
Answer:
Given the equation
x = 4y – 2
x – 2 = 4y
y = x/4 + 1/2

Question 4.
3y = -6x + 1
Answer:
Given the equation
3y = -6x + 1
y = -2x + 1/3

Question 5.
1 + y = –\(\frac{4}{5}\)x – 2
Answer:
Given the equation
1 + y = –\(\frac{4}{5}\)x – 2
y = –\(\frac{4}{5}\)x – 3

Question 6.
2.5y = 5x – 5
Answer:
Given the equation
2.5y = 5x – 5
y = 2x – 2

Question 7.
1.3y + 5.2 = -3.9x
Answer:
Given the equation
1.3y + 5.2 = -3.9x
1.3y = -3.9x – 5.2
y = -3x – 4

Question 8.
y – \(\frac{2}{3}\)x = -6
Answer:
Given the equation
y – \(\frac{2}{3}\)x = -6
y = \(\frac{2}{3}\)x -6

Concepts, Skills, &Problem Solving

GRAPHING A LINEAR EQUATION Graph the equation. (See Exploration 1, p. 161.)
Question 9.
The graph of y = 3.5x is translated up 2 units.
Answer:
Given the equation
y = 3.5x
The line obtained by translating the graph of the line y = 3.5x up 2 units has the same slope (3.5) and y – intercept 2 units greater, which means b = 0 + 2 = 2

Question 10.
The graph of y = -5x is translated down 3 units.
Answer:
y = -5x
The line obtained by translating the graph of the line y = -5x down 3 units has the same slope and the y – intercept 3 units smaller, which means b = 0 – 3 = -3

MATCHING EQUATIONS AND GRAPHS Match the equation with its graph. Identify the slope and the y-intercept.
Question 11.
y = 2x + 1
Answer:
Given the eqation
y = 2x + 1
slope = 2 and y – intercept = 1

Question 12.
y = \(\frac{1}{3}\)x – 2
Answer:
slope = 1/3 and y – intercept = -2

Question 13.
y = –\(\frac{2}{3}\)x + 1
Answer:

Answer:
Slope = -2/3 and y – intercept = 1
The graph which passes through the point (0, 1) and has a negative slope is the matching graph of the given equation.

IDENTIFYING SLOPES AND y-INTERCEPTS Find the slope and the y-intercept of the graph of the linear equation.
Question 14.
y = 4x – 5
Answer:
y = mx + b
slope = 4 and y — intercept = -5

Question 15.
y = -7x + 12
Answer:
y = -7x + 12
y = mx + b
slpoe = -7 and y – intercept = 12

Question 16.
y = –\(\frac{4}{5}\)x – 2
Answer:
y = mx + b
slope = -4/5
y – intercept = -2

Question 17.
y = 2.25x + 3
Answer:
y = mx + b
slope = 2.25 and y – intercept = 3

Question 18.
y + 1 = \(\frac{4}{3}\)x
Answer:
y = mx + b
y + 1 = \(\frac{4}{3}\)x
y = \(\frac{4}{3}\)x – 1
slope = \(\frac{4}{3}\), y – intercept = -1

Question 19.
y – 6 = \(\frac{3}{5}\)x
Answer:
y = mx + b
y – 6 = \(\frac{3}{5}\)x
y = \(\frac{3}{5}\)x + 6
slope = 3/8 and y – intercept = 6

Question 20.
y – 3.5 = -2x
Answer:
y = mx + b
y – 3.5 = -2x
y = -2x + 3.5
slope = -2 and y – intercept = 3.5

Question 21.
y = -5 – \(\frac{1}{2}\)x
Answer:
y = mx + b
y = -5 – \(\frac{1}{2}\)x
y =- \(\frac{1}{2}\)x – 5
slope = – \(\frac{1}{2}\) and y – intercept = -5

Question 22.
y = 11 + 1.5x
Answer:
y = mx + b
y = 1.5x + 11
slope = 1.5 and y – intercept = 11

Question 23.
YOU BE THE TEACHER
Your friend finds the slope and y-intercept of the graph of the equation y = 4x – 3. Is your friend correct? Explain your reasoning.

Answer:
y = 4x – 3
No my friend is not correct because the y – intercept is -3.

Question 24.
MODELING REAL LIFE
The number y of seasonal allergy shots available at a facility x days after receiving a shipment is represented by y = -15x + 375.
a. Graph the linear equation.
b. Interpret the slope and the y-intercept.
Answer:
y = -15x + 375
x = 0
y = -15(0) + 375 = 375
y = 0
0 = -15x + 375
15x = 375
x = 375/15
x = 25

The slope shows that the number of seasonal allergy shots decrease by 15 shots each day.
The y – intercept shows that the number of shots immediately after receiving a shipment is 375.

GRAPHING AN EQUATION Graph the linear equation. Identify the x-intercept.
Question 25.
y = x + 3
Answer:
Given the equation
y = x + 3
slope = 1 and y – intercept = 3
Slope = rise/run = 1/1
Plot the point that is 1 unit right and 1 unit up from (0, 3) = (1, 4)

So, the x – intercept is -3.

Question 26.
y = 4x – 8
Answer:
y = 4x – 8
Comparing the above equation with slope – intercept equation.
slope = 4 and y – intercept = -8
Slope = rise/run = 4/1 = 4
Plot the point that is 1 unit right and 4 unit up from (0, -8) = (1, -4)

Question 27.
y = -3x + 9
Answer:
y = -3x + 9
slope = -3 and y – intercept = 9
slope rise/run = -3/1 = -3

So, the intercept is 3.

Question 28.
y = -5x – 5
Answer:
y = -5x – 5
slope = -5 and y – intercept = -5
slope = rise/run = -5/1
Plot the point that is 1 unit right and 5 unit up from (0, -5) = (1, -10)

So, the x – intercept is -1.

Question 29.
y + 14 = -7x
Answer:
y + 14 = -7x
y = -7x – 14
slope = -7 and y – intercept = -14
Slope = rise/run = -7/1
Plot the point that is 1 unit right and 7 unit down from (0, -14) = (1, -21)

So, the x – intercept is -2.

Question 30.
y = 8 – 2x
Answer:
Given the equation
y = 8 – 2x
y = -2x + 8
slope = -2 and y – intercept = 8
slope = rise/run = -2/1
Plot the point 1 unit right and 2 units down from (0, 8) = (1, 6)

So, the x – intercept is 4.

Question 31.
PRECISION
You go to a harvest festival and pick apples.
a. Which equation represents the cost (in dollars) of going to the festival and picking x pounds of apples? Explain.

b. Graph the equation you chose in part(a).
Answer:
Picking a pound of apples costs $0.75, therefore x pounds cost 0.75 × x = 0.75x
y = 0.75x + 5

Question 32.
REASONING
Without graphing, identify the equations of the lines that are parallel. Explain your reasoning.

Answer:
The lines which area parallel are those having the same slope.
y = 2x + 4
y = 2x – 3
y = 2x + 1
y = 1/2x + 1
y = 1/2x + 2

Question 33.
PROBLEM SOLVING
A skydiver parachutes to the ground. The height y (in feet) of the skydiver after x seconds is y = -10x + 3000.

a. Graph the linear equation.
b. Interpret the slope, y-intercept, and x-intercept.
Answer:
y = -10x + 3000
x = 0
y = -10(0) + 3000 = 3000
y = 0
0 = -10 + 3000
10x = 3000
x = 3000/10 = 300

b. The slope shows that each second the skydiver descends 10 feet.
The y – intercept shows that the skydiver begins its dive from 3000 feet.
The x – intercept shows that he reaches the ground after 300 seconds.

Question 34.
DIG DEEPER!
Six friends create a website. The website earns money by selling banner ads. It costs $120 a month to operate the website.
a. A banner ad earns $0.005 per click. Write a linear equation that represents the monthly profit after paying operating costs.
b. Graph the equation in part(a). On the graph, label the number of clicks needed for the friends to start making a profit. Explain.
Answer:
y = 0.005x – 120
x = 0
y = 0.005(0) – 120
y = -120
y = 0
0 = 0.005x – 120
0.005x = 120
x = 24000

x > 24,000

Lesson 4.5 Graphing Linear Equations in Standard Form

EXPLORATION 1

Using Intercepts
Work with a partner. You spend $150 on fruit trays and vegetable trays for a party.

a. You buy x fruit trays and y vegetable trays. Complete the verbal model. Then use the verbal model to write an equation that relates x and y.

b. What is the greatest number of fruit trays that you can buy? vegetable trays? Can you use these numbers to graph your equation from part (a) in the coordinate plane? Explain.

c.Use a graph to determine the different combinations of fruit trays and vegetable trays that you can buy. Justify your answers algebraically.
d. You are given an extra $50 to spend. How does this affect the intercepts of your graph in part(c)? Explain your reasoning.
Answer:

Try It

Graph the linear equation.
Question 1.
x + y = -2
Answer:
Given the equation
y = mx + b
x + y = -2
y = -x – 2
Comparing the value of b and m from y = mx + b
m = -1 and b = -2
Plot y – intercept = (0, b) = (0, -2)
Slope = -1
run/rise = -1/1
Plot the point 1 unit down and 1 unit to the right = (1, -3)
Now plot the points and draw the graph

Question 2.
–\(\frac{1}{2}\)x + 2y = 6
Answer:
–\(\frac{1}{2}\)x + 2y = 6
2y = 6 + \(\frac{1}{2}\)x
y = 0.25x + 3
Comparing the value of b and m from y = mx + b
m = 0.25 and b = 3
Plot y – intercept = (0, b) = (0, 3)
Slope = 0.25
run/rise = 0.25/1
Plot the point 0.25 unit up and 1 unit to the right = (1, 3.25)
Now plot the points and draw the graph

Question 3.
–\(\frac{2}{3}\)x + y = 0
Answer:
–\(\frac{2}{3}\)x + y = 0
y = \(\frac{2}{3}\)x
Comparing the value of b and m from y = mx + b
m = \(\frac{2}{3}\) and b = 0
Plot y – intercept = (0, b) = (0, 0)
Slope =\(\frac{2}{3}\)
run/rise = \(\frac{2}{3}\)
Plot the point 0.25 unit up and 1 unit to the right = (3, 2)
Now plot the points and draw the graph

Question 4.
2x + y = 5
Answer:
2x + y = 5
y = -2x + 5
Comparing the value of b and m from y = mx + b
m = -2 and b = 5
Plot y – intercept = (0, b) = (0, 5)
Slope = -2
run/rise = \(\frac{-2}{1}\)
Plot the point 0.25 unit up and 1 unit to the right = (1, 3)
Now plot the points and draw the graph

Graph the linear equation using intercepts.
Question 5.
2x – y = 8
Answer:
y = 0
2x – y = 8
2x – 0 = 8
2x = 8
x = 4
The x – intercept is (4, 0)
Y – intercept :
x = 0
2x – y = 8
2(0) – y = 8
y = -8

Question 6.
x + 3y = 6
Answer:
X-intercept:
y = 0
x + 3y = 6
x + 3(0) = 6
x + 0 = 6
x = 6
The x – intercept is (6, 0)
Y – intercept:
x = 0
x + 3y = 6
0 + 3y = 6
y = 2
The y – intercept is (0, 2)

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

STRUCTURE Determine whether the equation is in standard form. If not, rewrite the equation in standard form.
Question 7.
y = x – 6
Answer:
y = x – 6
The standard form of equation is: Ax + By = C
The given equation is not in the standard form.
y = x – 6
x – y = 6

Question 8.
y – \(\frac{1}{6}\)x + 5 = 0
Answer:
The standard form of equation is: Ax + By = C
The given equation is not in the standard form.
y – \(\frac{1}{6}\)x + 5 = 0
\(\frac{1}{6}\)x – y = 5

Question 9.
4x + y = 5
Answer:
The standard form of equation is: Ax + By = C
The given equation is in the form of the standard form.

Question 10.
WRITING
Describe two ways to graph the equation 4x + 2y = 6.
Answer:
The two ways to graph the equation:
1. Graph the equation using standard form
2. Graph the equation using intercept.

GRAPHING A LINEAR EQUATION Graph the linear equation.
Question 11.
4x + y = 5
Answer:
Given the equation
4x + y = 5
y = -4x + 5
Comparing the value of b and m from y = mx + b
m = -4 and b = 5
Plot y – intercept = (0, b) = (0, 5)
Slope = -4
run/rise = \(\frac{-4}{1}\)
Plot the point 4 unit down and 1 unit to the right = (1, 1)
Now plot the points and draw the graph

Question 12.
\(\frac{1}{3}\)x + 2y = 8
Answer:
X – intercept:
y = 0
\(\frac{1}{3}\)x + 2y = 8
\(\frac{1}{3}\)x + 2(0) = 8
\(\frac{1}{3}\)x = 8
x = 24
The x – intercept is (24, 0)
Y – intercept:
x = 0
\(\frac{1}{3}\)x + 2y = 8
\(\frac{1}{3}\)(0) + 2y = 8
2y = 8
y = 4
The y – intercept is (0, 4)

Question 13.
5x – y = 10
Answer:
X – intercept:
y = 0
5x – 0 = 10
5x = 10
x = 2
The x-intercept is (2, 0)
Y – intercept:
x = 0
5x – y = 10
5(0) – y = 10
-y = 10
y = -10
The y – intercept is (0, -10)

Question 14.
x – 3y = 9
Answer:
X – intercept:
y = 0
x – 3(0) = 9
x = 9
The x – intercept is (9, 0)
Y – intercept:
x = 0
0 – 3y = 9
-3y = 9
y = -3
The y – intercept is (0, -3)

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 15.
You have $30 to spend on paint and clay. The equation 2x + 6y = 30 represents this situation, where x is the number of paint bottles and y is the number of tubs of clay. Graph the equation. Interpret the intercepts. How many bottles of paint can you buy if you buy 3 tubs of clay? Justify your answer.
Answer:
Given,
You have $30 to spend on paint and clay.
The equation 2x + 6y = 30 represents this situation,
where x is the number of paint bottles and y is the number of tubs of clay.
X – intercept:
y = 0
2x + 6y = 30
2x + 6(0) = 30
2x = 30
x = 15
The x – intercept is (15, 0)
Y – intercept:
x = 0
2x + 6y = 30
2(0) + 6y = 30
6y = 30
y = 5
The y – intercept is (0, 5)

From the graph, I can buy 6 bottles of point if I buy 3 tubs of clay.

Question 16.
You complete two projects for a class in 60 minutes. The equation x + y = 60 represents this situation, where x is the time (in minutes) you spend assembling a birdhouse and y is the time (in minutes) you spend writing a paper.

a. Graph the equation. Interpret the intercepts.

Answer:
x + y = 60
y = -x + 60

b. You spend twice as much time assembling the birdhouse as you do writing the paper. How much time do you spend writing the paper? Justify your answer.
Answer:
We are given,
y = 2x
2x = -x + 60
2x + x = 60
3x = 60
x = 20
y = 2 (20)
y = 40

Graphing Linear Equations in Standard Form Homework & Practice 4.5

Review & Refresh

Find the slope and the y-intercept of the graph of the linear equation.
Question 1.
y = x – 1
Answer:
y = mx + b
Slope = -1 and y – intercept = -1

Question 2.
y = -2x + 1
Answer:
y = -2x + 1
y = mx + b
Slope = -2 and y – intercept = 1

Question 3.
y = \(\frac{8}{9}\)x – 8
Answer:
y = \(\frac{8}{9}\)x – 8
y = mx + b
Slope = \(\frac{8}{9}\) and y – intercept = -8

Tell whether the blue figure is a reflection of the red figure.
Question 4.

Answer:
The blue figure is not a reflection of the red figure because, for example the reflection of the upper leg of the upper leg of the red triangle across the y-axis is the top vertex of the blue triangle, not a point.

Question 5.

Answer:
The blue figure is a reflection of the red figure because to each point in the red figure corresponds a symmetrical point in the blue figure.

Question 6.

Answer:
The blue figure is a reflection of the red figure because to each point in the red figure corresponds a symmetrical point in the blue figure.

Concepts, Skills, &Problem Solving

USING INTERCEPTS Define two variables for the verbal model. Write an equation in slope-intercept form that relates the variables. Graph the equation using intercepts. (See Exploration 1, p. 167.)
Question 7.

Answer:
x = amount of peaches (in pounds)
y = the amount of apples (in pounds)
2x + 1.5y = 15
y = 0 = 2x + 1.5(0) = 15
2x = 15
x = 7.5
x = 0
2(0) + 1.5y = 15
1.5y =15
y = 10

Question 8.

Answer:
x = the biked distance (in miles)
y = the walked distance (in miles)
y = 0
16x + 2(0) = 32
16x = 32
x = 2
x = 0
16(0) + 2y = 32
2y = 32
y = 16

REWRITING AN EQUATION Write the linear equation in slope-intercept form.
Question 9.
2x + y = 17
Answer:
Given the equation
2x + y = 17
y = 17 – 2x
y = -2x + 17

Question 10.
5x – y = \(\frac{1}{4}\)
Answer:
Given the equation
5x – y = \(\frac{1}{4}\)
-y = \(\frac{1}{4}\) – 5x
y = 5x – \(\frac{1}{4}\)

Question 11.
–\(\frac{1}{2}\)x + y = 10
Answer:
Given the equation
–\(\frac{1}{2}\)x + y = 10
y = \(\frac{1}{2}\)x + 10

GRAPHING AN EQUATION Graph the linear equation.
Question 12.
-18x + 9y = 72
Answer:
Given the equation
-18x + 9y = 72
X – intercept:
y = 0
-18x + 9(0) = 72
-18x = 72
x = -4
The x – intercept is (-4, 0)
Y – intercept:
x = 0
-18x + 9y = 72
-18(0) + 9y = 72
9y = 72
y = 8

Question 13.
16x – 4y = 2
Answer:
Given the equation
16x – 4y = 2
X – intercept:
y = 0
16x – 4y = 2
16x – 4(0) = 2
16x = 2
x = 0.125
The X – intercept is (0.125, 0)
Y – intercept:
x = 0
16(0) – 4y = 2
-4y = 2
y = -2

Question 14.
\(\frac{1}{4}\)x + \(\frac{3}{4}\)y = 1
Answer:
Given the equation
\(\frac{1}{4}\)x + \(\frac{3}{4}\)y = 1
x + 3y = 4
y = 0
x + 3(0) = 4
x = 4
x = 0
0 + 3y = 4
3y = 4
y = 4/3

MATCHING Match the equation with its graph.
Question 15.
15x – 12y = 60
Answer:
y = 0
15x – 12(0) = 60
15x = 60
x = 60/15
x = 4
x = 0
15(0) – 12y = 60
-12y = 60
y = -5
The graph having the x – intercept 4 and y – intercept -5

Question 16.
5x + 4y = 20
Answer:
Given the linear equation
5x + 4y = 20
y = 0
5x + 4(0) = 20
5x = 20
x = 4
x = 0
5(0) + 4y = 20
4y = 20
y = 5

Question 17.
10x + 8y = -40
Answer:

10x + 8y = -40
y = 0
10x + 8(0) = -40
10x = -40
x = -4
x = 0
10(0) + 8y = -40
8y = -40
y = -5

Question 18
YOU BE THE TEACHER
Your friend finds the x-intercept of -2x + 3y = 12. Is your friend correct? Explain your reasoning.

Answer:
-2x + 3y = 12
y = 0
-2x + 3(0) = 12
-2x = 12
x = -6
Your friend is not correct because the x – intercept is the value of x corresponding to y = 0.
Your friend computed the y – intercept.

Question 19.
MODELING REAL LIFE
A charm bracelet costs $65, plus $25 for each charm. The equation -25x + y = 65 represents the cost y (in dollars) of the bracelet, where x is the number of charms.
a. Graph the equation.
b. How much does a bracelet with three charms cost?
Answer:

y = 25x + 65
Substitute the value of x in the equation
y = 25(3) + 65
y = 75 + 65
y = 140

USING INTERCEPTS TO GRAPH Graph the linear equation using intercepts.
Question 20.
3x – 4y = -12
Answer:
Given the equation
3x – 4y = -12
3x – 4(0) = -12
3x = -12
x = -4
The x – intercept is (-4, 0)
Y – intercept:
x = 0
3(0) – 4y = -12
-4y = -12
y = 3
The y – intercept is (0, 3)

Question 21.
2x + y = 8
Answer:
X – intercept:
y = 0
2x + y = 8
2x + 0 = 8
2x = 8
x = 4
The x – intercept is (4, 0)
Y – intercept:
x = 0
2x + y = 8
2(0) + y = 8
y = 8
The y – intercept is (0, 8)

Question 22.
\(\frac{1}{3}\)x – \(\frac{1}{6}\)y = –\(\frac{2}{3}\)
Answer:
X – intercept:
y = 0
\(\frac{1}{3}\)x – \(\frac{1}{6}\)(0) = –\(\frac{2}{3}\)
\(\frac{1}{3}\)x = –\(\frac{2}{3}\)
x = -2
The x – intercept is (-2, 0)
Y – intercept:
x = 0
\(\frac{1}{3}\)(0) – \(\frac{1}{6}\)y = –\(\frac{2}{3}\)
y = 4
The y – intercept is (0, 4)

Question 23.
MODELING REAL LIFE
Your cousin has $90 to spend on video games and movies. The equation 30x + 15y = 90 represents this situation, where x is the number of video games purchased and y is the number of movies purchased. Graph the equation. Interpret the intercepts.
Answer:
30x + 15y = 90
x = 0
30(0) + 15y = 90
15y = 90
y = 6
y = 0
30x + 15(0) = 90
30x = 90
x = 3

The x – intercept shows that 3 video games are purchased when no movies are purchased.
The y – intercept shows that 6 movies are purchased when no video games are purchased.

Question 24.
PROBLEM SOLVING
A group of friends go scuba diving. They rent a boat for x days and scuba gear for y people, represented by the equation 250x + 50y = 1000.

a. Graph the equation and interpret the intercepts.
b. How many friends can go scuba diving if they rent the boat for 1 day? 2 days?
c. How much money is spent in total?
Answer:
250x + 50y = 1000
x = 0
250(0) + 50y = 1000
50y = 1000
y = 20
when y = 0
250x + 50(0) = 1000
250x = 1000
x = 4

b.
250(1) + 50y = 1000
250 + 50y = 1000
50y = 1000 – 250
50y = 750
y = 15
when x = 2
250(2) + 50y = 1000
500 + 50y = 1000
50y = 1000 – 500
50y = 500
y = 500/50
y = 10

Question 25.
DIG DEEPER!
You work at a restaurant as a host and a server. You earn $9.45 for each hour you work as a host and $3.78 for each hour you work as a server.

a. Write an equation in standard form that models your earnings.
b. Graph the equation.
Answer:
You earn $9.45 for each hour you work as a host and $3.78 for each hour you work as a server.
Number of hours worked as host + $3.78.
Number of hours worked as server = $113.40
9.45x + 3.78y = 113.40
x = 0
9.45(0) + 3.78y = 113.40
3.78y = 113.40
y = 30
when y = 0
9.45x + 3.78(0) = 113.40
9.45x = 113.40
x = 12

Question 26.
LOGIC
Does the graph of every linear equation have an x-intercept? Justify your reasoning.
Answer:
y = mx + b
y = 0
0 = mx + b
mx = -b
x = -b/m for m ≠ 0
If m = 0 the equation has no solution. Therefore the equation y = b has no x – intercept.

Question 27.
CRITICAL THINKING
For a house call, a veterinarian charges $70, plus $40 per hour.

a. Write an equation that represents the total fee y (in dollars) the veterinarian charges for a visit lasting x hours.

b. Find the x-intercept. Does this value make sense in this context? Explain your reasoning.
c. Graph the equation.
Answer:
Total fee = fixed charge + number of hours . cost per hour
y = 70 + 40x
y = 0
0 = 70 + 40x
-70 = 40x
x = -1.75
x = 0
y = 70 + 40(0)
y = 70

Lesson 4.6 Writing Equations in Slope-Intercept Form

EXPLORATION 1

Writing Equations of Lines
Work with a partner.For each part, answer the following questions.

• What are the slopes and the y-intercepts of the lines?
• What are equations that represent the lines?
• What do the lines have in common?

Answer:

EXPLORATION 2

Interpreting the Slope and the y-Intercept
Work with a partner. The graph represents the distance y (in miles) of a car from Phoenix after t hours of a trip.

a. Find the slope and the y-intercept of the line. What do they represent in this situation?
b. Write an equation that represents the graph.
c. How can you determine the distance of the car from Phoenix after 11 hours?
Answer:

Try It

Write an equation in slope-intercept form of the line that passes through the given points.
Question 1.

Answer:
m = (y2 – y1)/(x2 – x1)
= (4 – 2)/(1 – 0)
= 2/1
= 2
Because the line crosses the y – axis at (0, 2)
y = mx + b
y = 2x + 2

Question 2.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-1 – 3)/(0 – (-3))
= -4/3
Because y = -1 when x = 0, the y – intercept is -1
y = mx + b
y = -4/3 x – 1

Write an equation of the line that passes through the given points.
Question 3.

Answer:
m = (y2 – y1)/(x2 – x1)
= (5 – 5)/(0 – (-4))
= 0/4
Because y = 5 when x = 0, the y – intercept is 5
y = mx + b
y = (0)x + 5
y = 5

Question 4.

Answer:
m = (y2 – y1)/(x2 – x1)
= (1 – 1)/(3 – 0)
= 0/3
= 0
Because the line crosses the y – axis at (0, 1) the y – intercept is 1
y = mx + b
y = (0)x + 1
y = 1

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING EQUATIONS IN SLOPE-INTERCEPT FORM Write an equation in slope-intercept form of the line that passes through the given points.
Question 5.

Answer:
m = (y2 – y1)/(x2 – x1)
= (5 – 2)/(1 – 0)
= 3/1
= 3
Because y = 2 when x = 0, the y – intercept is 2
y = mx + b
y = (3)x + 2
y = 3x + 2

Question 6.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-1 – 5)/(1 – (-1))
= -6/2
= -3
Because the line crosses the y – axis at (0, 2) the y – intercept is 2
y = mx + b
y = -3x + 2

Question 7.
WRITING AN EQUATION
Write an equation of the line that passes through (0, -5) and (2, -5).
Answer:
m = (y2 – y1)/(x2 – x1)
= (-5 – (-5))/(2 – 0)
= 0/2
= 0
Because y = -5 when x = 0, the y – intercept is -5
y = mx + b
y = (0)x + -5
y = -5

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 8.
You load boxes onto an empty truck at a constant rate. After 3 hours, there are 100 boxes on the truck. How much longer do you work if you load a total of 120 boxes? Justify your answer.
Answer:
Let x be the number of hours you work if you load a total of 120 boxes.
100/3 = 120/x
100x = 3 × 120
x = 360/100
x = 3.6 hours
3.6 – 3 = 0.6 hours

Question 9.
The table shows the amounts (in tons) of waste left in a landfill after x months of waste relocation. Interpret the slope and the y-intercept of the line that passes through the given points. How many months does it take to empty the landfill? Justify your answer.

Answer:
m = (12 – 15)/ (6 – 0)
m = -3/6
m = -0.5
b = 15
The y – intercept shows that there are 150 tons of waste in the beginning.
y = -0.5x + 15
y = 0
0 = -0.5x + 15
x = 30
So the ladfill will be emptied after 30 months.

Question 10.
DIG DEEPER!
A lifetime subscription to a website costs $250. A monthly subscription to the website costs $10 to join and $15 per month. Write equations to represent the costs of each plan. If you want to be a member for one year, which plan is less expensive? Explain.
Answer:
Given,
A lifetime subscription to a website costs $250. A monthly subscription to the website costs $10 to join and $15 per month.
Total cost for plan 1 = the lifetime subscription
y = 250
Total cost for Plan 2 = Fixed tax + Number of months . monthly cost
y = 10 + 15x
Plan 1: y = 250
Plan 2: y = 10 + 15(12) = 190
As 190 < 250, plan 1 is less expensive.

Writing Equations in Slope-Intercept Form Homework & Practice 4.6

Review & Refresh

Write the linear equation in slope-intercept form.
Question 1.
4x + y = 1
Answer:
Given the equation
4x + y = 1
y = -4x + 1

Question 2.
x – y = \(\frac{1}{5}\)
Answer:
Given the equation
x – y = \(\frac{1}{5}\)
x – \(\frac{1}{5}\) = y

Question 3.
–\(\frac{2}{3}\)x + 2y = -7
Answer:
Given the equation
–\(\frac{2}{3}\)x + 2y = -7
2y = -7 + \(\frac{2}{3}\)x
y = \(\frac{1}{3}\)x – \(\frac{7}{2}\)

Plot the ordered pair in a coordinate plane.
Question 4.
(1, 4)
Answer:

Question 5.
(-1, -2)
Answer:

Question 6.
(0, 1)
Answer:

Question 7.
(2, 7)
Answer:

Concepts, Skills, & Problem Solving

INTERPRETING THE SLOPE AND THE y-INTERCEPT The graph y represents the cost (in dollars) to open an online gaming account and buy x games. (See Exploration 2, p. 173.)

Question 8.
Find the slope and the y-intercept of the line. What do they represent in this situation?
Answer:
(0, 15), (3, 45)
m = (45 – 15)/(3 – 0)
m = 30/3 10
Thus the slope of the line is m – 3.
b = 15
The slope represents the cost of one game, while the y – intercept is the cost of opening the gaming account.

Question 9.
Write an equation that represents the graph.
Answer:
m = 10
b = 15
y = mx + b
y = 10x + 15

Question 10.
How can you determine the total cost of opening an account and buying 6 games?
Answer:
y = 10x + 15
y = 10(6) + 15
y = 60 + 15
y = 75

WRITING EQUATIONS IN SLOPE-INTERCEPT FORM Write an equation in slope-intercept form of the line that passes through the given points.
Question 11.

Answer:
m = (y2 – y1)/(x2 – x1)
= (4 – 3)/(0 – (-1))
= 1/1
= 1
Because the line crosses the y – axis at (0, 4) the y – intercept is 4
y = mx + b
y = (1)x + 4
y = x + 4

Question 12.

Answer:
m = (y2 – y1)/(x2 – x1)
= (6 – 0)/(-3 – 0)
= 6/-3
= -2
Because the line crosses the y – axis at (0, 2) the y – intercept is 2
y = mx + b
y = -2x + 0
y = -2x

Question 13.

Answer:
m = (y2 – y1)/(x2 – x1)
= (2 – 1)/(4 – 0)
= 1/4
Because the line crosses the y – axis at (0, 1) the y – intercept is 1
y = mx + b
y = 1/4 x + 1

Question 14.

Answer:
m = (y2 – y1)/(x2 – x1)
= (1 – 2)/(0 – (-2))
= -1/2
Because y = 1 when x = 0, the y – intercept is 1
y = mx + b
y = -1/2 x + 2

Question 15.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-3 – (-4))/(0 – (-3))
= 1/3
Because y = -3 when x = 0, the y – intercept is -3
y = mx + b
y = 1/3 x – 3

Question 16.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-1 -4)/(0 – (-2))
= -5/2
Because y = -1 when x = 0, the y – intercept is -1
y = mx + b
y = -5/2 x – 1

WRITING EQUATIONS Write an equation of the line that passes through the given points.
Question 17.
(-1, 4), (0, 2)
Answer:
m = (y2 – y1)/(x2 – x1)
= (2 – 4)/(0 – (-1))
= -2/1
= -2
Because y = 2 when x = 0, the y – intercept is 2
y = mx + b
y = -2x + 2

Question 18.
(-1, 0), (0, 0)
Answer:
m = (y2 – y1)/(x2 – x1)
= (0 – 0)/(0 – (-1))
= 0/1
= 0
Because y = 0 when x = 0, the y – intercept is 0
y = mx + b
y = 0

Question 19.
(0, 4), (0, -3)
Answer:
Both points belong to the y-axis. Therefore the equation of the line passing through them is
x = 0

Question 20.
YOU BE THE TEACHER
Your friend writes an equation of the line shown. Is your friend correct? Explain your reasoning.

Answer:
Because in the given graph, y = -2 when x = 0, so the y – intercept is -2. The equation of the line should be: y = 1/2 x – 2
No my friend is NOT correct.

Question 21.
MODELING REAL LIFE
A boa constrictor is 18 inches long at birth and grows 8 inches per year. Write an equation in slope y-intercept form that represents the length (in feet) of a boa constrictor that is x years old.

Answer:
Given,
A boa constrictor is 18 inches long at birth and grows 8 inches per year.
Length after x years = birth length + number of years . Growth per year
y = 18 + 8x
y = 8x + 18
Convert it into feet
y = 2/3 x + 3/2

Question 22.
MODELING REAL LIFE
The table shows the speeds y (in miles per hour) of a car after x seconds of braking. Write an equation of the line that passes through the points in the table. Interpret the slope and the y-intercept of the line.

Answer:
m = (y2 – y1)/(x2 – x1)
= (60 – 70)/(1 – 0)
= -10/1
= -10
Because y = 70 when x = 0, the y – intercept is 70
y = mx + b
y = -10x + 70
Slope = -10 represents the decrease in the speed of the car each seconds after breaking.
The y – intercept of 70 represents the initial speed of the car.

Question 23.
MODELING REAL LIFE
A dentist charges a flat fee for an office visit, plus an additional fee for every tooth removed. The graph shows the total cost y (in dollars) for a patient when the dentist removes x teeth. Interpret the slope and the y-intercept.

Answer:
(2, 500), (4, 900)
m = (900 – 500)/(4 – 2)
m = 400/2
m = 200
y = mx + b
500 = 200(2) + b
500 = 400 + b
b = 500 – 400
b = 100
The slope shows that the amount charged for each removed tooth is $200.
The y – intercept shows that the flat fee for an office visit is $100.

Question 24.
MODELING REAL LIFE
One of your friends gives you $10 for a charity walkathon. Another friend gives you an amount per mile. After 5 miles, you have raised $13.50 total. Write an equation that represents the amount y of money you have raised after x miles.
Answer:
Given,
One of your friends gives you $10 for a charity walkathon.
Another friend gives you an amount per mile. After 5 miles, you have raised $13.50 total.
y = mx + b
b = 10
13.50 = 5m + 10
13.50 – 10 = 5m
3.50 = 5m
m = 3.50/5
m = 0.7
y = 0.7x + 10

Question 25.
PROBLEM SOLVING
You have 500 sheets of notebook paper. After 1 week, you have 72% of the sheets left. You use the same number of sheets each week. Write an equation that represents the number y of sheets remaining after x weeks.
Answer:
y = mx + b
500 – 0.72 × 500 = 500 – 360 = 140 sheets
m = -140
b = 500
y = -140x + 500

Question 26.
DIG DEEPER!
The palm tree on the left is 10 years old. The palm tree on the right is 8 years old. The trees grow at the same rate.

a. Estimate the height y (in feet) of each tree.
b. Plot the two points (x, y), where x is the age of each tree and y is the height of each tree.
c. What is the rate of growth of the trees?
d. Write an equation that represents the height of a palm tree in terms of its age.
Answer:
a. estimate
left: 18
right: 12
plot y = 1.8x

Lesson 4.7 Writing Equations in Point-Slope Form

EXPLORATION 1

Deriving an Equation
Work with a partner. Let (x₁, y₁) represent a specific point on a line. Let (x, y) represent any other point on the line.

a. Write an equation that represents the slope m of the line. Explain your reasoning.
b. Multiply each side of your equation in part(a) by the expression in the denominator. What does the resulting equation represent? Explain your reasoning.
Answer:

EXPLORATION 2

Writing an Equation
Work with a partner.
For 4 months, you saved $25 a month. You now have $175 in your savings account.

a. Draw a graph that shows the balance in your account after t months.
b.Use your result from Exploration 1 to write an equation that represents the balance A after t months.
Answer:

Try It
Write an equation in point -slope form of the line that passes through the given point and has the given slope.
Question 1.
(1, 2); m = -4
Answer:
y – y1 = m(x – x1)
y – 2 = -4(x – (1))
y – 2 = -4(x – 1)

Question 2.
(7, 0); m = 1
Answer:
y – y1 = m(x – x1)
y – 0 = 1(x – (7))
y – 0 = 1(x – 7)

Question 3.
(-8, -5); m = –\(\frac{3}{4}\)
Answer:
y – y1 = m(x – x1)
y – (-5) = –\(\frac{3}{4}\)(x – (-8))
y + 5 = –\(\frac{3}{4}\)(x + 8)

Write an equation in slope-intercept form of the line that passes through the given points.
Question 4.
(-2, 1), (3, -4)
Answer:
Slope(m) = (-4 – 1)/(3 – (-2))
= -5/5
m = -1
y – y1 = m(x – x1)
y – 1 = -1(x – (-2))
y – 1 = -1(x + 2)
y – 1 = -x – 2
y = -x – 1

Question 5.

Answer:
Slope(m) = (3 – 5)/(-3 – (-5))
= -2/2
m = -1
y – y1 = m(x – x1)
y – 1 = -1(x – (-1))
y – 1 = -1(x + 1)
y – 1 = -x – 1
y = -x – 1 + 1
y = -x

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

WRITING AN EQUATION Write an equation in point-slope form of the line that passes through the given point and has the given slope.
Question 6.
(2, 0); m = 1
Answer:
y – y1 = m(x – x1)
y – 0 = 1(x – (2))
y – 0 = 1(x – 2)

Question 7.
(-3, -1); m = –\(\frac{1}{3}\)
Answer:
y – y1 = m(x – x1)
y – (-1) = –\(\frac{1}{3}\)(x – (-3))
y + 1 = –\(\frac{1}{3}\)(x + 3)

Question 8.
(5, 4); m = 3
Answer:
y – y1 = m(x – x1)
y – 4 = 3(x – (5))
y – 4 = 3(x – 5)

Question 9.
WRITING AN EQUATION
Write an equation of the line that passes through the points given in the table.

Answer:
Slope(m) = (-2 – 1)/(5 – 3)
= -3/2
m = -1
y – y1 = m(x – x1)
y – (-5) = -3/2(x – 7)
y + 5 = -3/2(x – 7)
y + 5 = -3/2 x + 21/2
y = -3/2 x + 11/2

Question 10.
DIFFERENT WORDS, SAME QUESTION
Which is different? Sketch “both” graphs.

Answer:
y – 7 = 4x – 4
y = 4x + -4 + 7
y = 4x + 3
Graph line passes through the points (4, 5) and (5, 9)

Self-Assessment for Problem Solving
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.
A writer finishes a project that a coworker started at a rate of 3 pages per hour. After 3 hours,25% of the project is complete.
a. The project is 200 pages long. Write and graph an equation for the total number y of pages that have been finished after the writer works for x hours.
b. The writer has a total of 45 hours to finish the project. Will the writer meet the deadline? Explain your reasoning.
Answer:
m = 3
y = 3x + b
b + 9 = 25%(200)
b + 9 = 0.25(200)
b + 9 = 50
b = 50 – 9
b = 41
y = 3x + 41

y = 3x + 41
y = 3(45) + 41 = 176 pages
As 176 < 200, the writer will not meet the deadline.

Question 12.
DIG DEEPER!
You and your friend begin to run along a path at different constant speeds.After 1 minute,your friend is 45 meters ahead of you. After 3 minutes, your friend is 105 meters ahead of you.
a. Write and graph an equation for the distance y (in meters) your friend is ahead of you after x minutes. Justify your answer.

Answer:
y = mx + b
45 = m + b
105 = 3m + b
105 – 45 = (3m + b) – (m + b)
60 = 2m
m = 30
45 = 30 + b
b = 45 – 30
b = 15
y = 30x + 15

b. Did you and your friend start running from the same spot? Explain your reasoning.
Answer:
The distance between you and your friend in the initial moment is b = 15 meters. So you are ahead your friend by 15 meters at the starting point.

Writing Equations in Point-Slope Form Homework & Practice 4.7

Review & Refresh

Write an equation in slope-intercept form of the line that passes through the given points.
Question 1.

Answer:
Slope(m) = (5 – 4)/(0 – (-2))
= 1/2
m = 1/2
Because y = 5 when x = 0, the y – intercept is 5.
y = mx + b
y = 1/2 x + 5

Question 2.

Answer:
Slope(m) = (5 – (-1))/(2 – (-2))
= (5 + 1)/(2 + 2)
m = 6/4
m = 3/2
From the graph, the line crosses the y – axis at (0, 2)
y = mx + b
y = 3/2 x + 2

Solve the equation. Check your solution, if possible.
Question 3.
2x + 3 = 2x
Answer:
Given the equation
2x + 3 = 2x
3 = 2x – 2x
3 ≠ 0

Question 4.
6x – 7 = 1 – 3x
Answer:
Given the equation
6x – 7 = 1 – 3x
6x + 3x = 1 + 7
9x = 8
x = 8/3

Question 5.
0.1x – 1 = 1.2x – 5.4
Answer:
Given the equation
0.1x – 1 = 1.2x – 5.4
0.1x – 1.2x = 1 – 5.4
-1.1x = -4.4
x = 4

Concepts, Skills, &Problem Solving

WRITING AN EQUATION The value of a new car decreases $4000 each year. After 3 years, the car is worth $18,000. (See Exploration 2, p. 179.)
Question 6.
Draw a graph that shows the value of the car after t years.
Answer:

Question 7.
Write an equation that represents the value V of the car after t years.
Answer:
y = -4000t + b
where b is the original price
18,000 = -4000(3) + b
18,000 + 12,000 = b
b = 30,000
y = -4000t + 30,000

WRITING AN EQUATION Write an equation in point-slope form of the line that passes through the given point and has the given slope.
Question 8.
(3, 0); m = –\(\frac{2}{3}\)
Answer:
y – y1 = m(x – x1)
y – (0) = -2/3(x – 3)
y – 0 = -2/3(x – 3)

Question 9.
(4, 8); m = \(\frac{3}{4}\)
Answer:
y – y1 = m(x – x1)
y – (8) = 3/4(x – 4)
y – 8 = 3/4(x – 4)

Question 10.
(1, -3); m = 4
Answer:
y – y1 = m(x – x1)
y – (-3) = 4(x – 1)
y + 3 = 4(x – 1)

Question 11.
(7, -5); m = –\(\frac{1}{7}\)
Answer:
y – y1 = m(x – x1)
y – (-5) = –\(\frac{1}{7}\)(x – 7)
y + 5 = –\(\frac{1}{7}\)(x – 7)

Question 12.
(3, 3); m = \(\frac{5}{3}\)
Answer:
y – y1 = m(x – x1)
y – (3) = \(\frac{5}{3}\)(x – 3)
y – 3 = \(\frac{5}{3}\)(x – 3)

Question 13.
(-1, -4); m = -2
Answer:
y – y1 = m(x – x1)
y – (-4) = -2(x – (-1))
y + 4 = -2(x + 1)

WRITING AN EQUATION Write an equation in slope-intercept form of the line that passes through the given points.
Question 14.
(-1, -1), (1, 5)
Answer:
Slope(m) = (5 – (-1))/(2 – (-1))
= (5 + 1)/(1 + 1)
m = 6/2
m = 3
y – y1 = m(x – x1)
y – (5) = 3(x – (1))
y – 5 = 3x – 3
y = 3x + 2

Question 15.
(2, 4), (3, 6)
Answer:
Slope(m) = (6 – 4)/(3 – 2)
m = 2/1
m = 2
y – y1 = m(x – x1)
y – (4) = 2(x – (2))
y – 4 = 2x – 4
y = 2x

Question 16.
(-2, 3), (2, 7)
Answer:
Slope(m) = (7 – (3))/(2 – (-2))
= (7 – 3)/(2 + 2)
m = 4/4
m = 1
y – y1 = m(x – x1)
y – (3) = 1(x – (-2))
y – 3 = x + 2
y = x + 5

Question 17.
(4, 1), (8, 2)
Answer:
Slope(m) = (2 – (1))/(8 – (4))
= (2 – 1)/(8 – 4)
m = 1/4
y – y1 = m(x – x1)
y – (1) = 1/4(x – (4))
y – 1 = 1/4 x – 1
y = 1/4 x

Question 18.
(-9, 5), (-3, 3)
Answer:
Slope(m) = (3 – (5))/(-3 – (-9))
= (3 – 5)/(-3 + 9)
m = -2/6
m = -1/3
y – y1 = m(x – x1)
y – (3) = -1/3(x + 3)
y – 3 = -1/3 x – 1
y = -1/3 x + 2

Question 19.
(1, 2), (-2, -1)
Answer:
Slope(m) = (2 – (1))/(8 – (4))
= (-1 – 2)/(-2 – 1)
m = -3/-3
m = 1
y – y1 = m(x – x1)
y – (2) = 1(x – (1))
y – 2 = x – 1
y = x + 1

Question 20.
MODELING REAL LIFE
At 0° C, the volume of a gas is 22 liters. For each degree the temperature T (in degrees Celsius) increases, the volume V (in liters) of the gas increases by \(\frac{2}{25}\). Write an equation that represents the volume of the gas in terms of the temperature.
Answer:
The equation modeling the situation has the form:
V = mT + b
m = 2/25
22 = 2/25(0) + b
b = 22
V = 2/25 T + 22

WRITING AN EQUATION Write an equation of the line that passes through the given points in any form. Explain your choice of form.
Question 21.

Answer:
m = (y2 – y1)/(x2 – x1)
= (2.5 – 1.5)/(0 – (-1))
= 1/1
= 1
Because the line crosses the y – axis at (0, 2.5), the y – intercept is 2.5
y = mx + b
y = (1)x + 2.5
y = x + 2.5

Question 22.

Answer:
m = (y2 – y1)/(x2 – x1)
= (3.5 – 1.5)/(2 – (1))
= 2/1
= 2
y – y1 = m(x – x1)
y – (1.5) = 2(x – (1))
y – 1.5 = 2x – 2
y = 2x – 0.5

Question 23.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-1.5 – 4.5)/(1 – (-1))
= -6/2
= -3
y – y1 = m(x – x1)
y – (-1.5) = -3(x – (1))
y + 1.5 = -3x + 3
y = -3x + 1.5

Question 24.

Answer:
m = (y2 – y1)/(x2 – x1)
= (-0.5 – 3.5)/(1 – (-1))
= -4/2
= -2
y – y1 = m(x – x1)
y – (-0.5) = -2(x – (1))
y + 0.5 = -2x – 2
y = -2x – 2.5

Question 25.

Answer:
m = (y2 – y1)/(x2 – x1)
= (1 – (-1))/(0 – (-3))
= (1 + 1)/(0 + 3)
= 2/3
Because y = 1 when x = 0, the y – intercept is 1.
y = mx + b
y = 2/3 x + 1

Question 26.

Answer:
m = (y2 – y1)/(x2 – x1)
= (4 – 6)/(-3 – (-7))
= -2/4
= -1/2
y – y1 = m(x – x1)
y – (2) = -1/2(x – (1))
y – 2 = -1/2x + 1/2
y = -1/2 x + 5/2

Question 27.
REASONING
Write an equation of the line that passes through the point (8, 2) and is parallel to the graph of the equation y = 4x – 3.
Answer:
y = 4x – 3
Comparing the given equation with y = mx + b, we get
m = 4
y – y1 = m(x – x1)
y – 2 = 4(x – 8)
y – 2 = 4x – 32
y = 4x – 32 + 2
y = 4x – 30

Question 28.
MODELING REAL LIFE
The table shows the amount y (in fluid ounces) of carpet cleaner in a tank after x minutes of cleaning.

a. Write an equation that represents the amount of cleaner x in the tank after minutes.
b. How much cleaner is in the tank when the cleaning begins?
c. After how many minutes is the tank empty? Justify your answer.
Answer:

Question 29.
DIG DEEPER!
According to Dolbear’s law, you can predict the temperature T (in degrees Fahrenheit) by counting the number x of chirps made by a snowy tree cricket in 1 minute.When the temperature is 50°F, a cricket chirps 40 times in 1 minute. For each rise in temperature of 0.25°F, the cricket makes an additional chirp each minute.
a. You count 100 chirps in 1 minute. What is the temperature?
b. The temperature is 96°F.How many chirps do you expect the cricket to make? Justify your answer.
Answer:

Question 30.
PROBLEM SOLVING
The Leaning Tower of Pisa in Italy was built between 1173 and 1350.
a. Write an equation that represents the yellow line.
b. The tower is 56 meters tall. How far from the center is the top of the tower? Justify your answer.

Answer:

Graphing and Writing Linear Equations Connecting Concepts

Using the Problem-Solving Plan
Question 1.
Every item in a retail store is on sale for 40% off. Write and graph an equation that represents the sale price of an item that has an original price of x dollars.

Understand the problem.
You know the percent discount of items in a retail store.You are asked to write and graph an equation that represents the sale price of an item that has an original price of x dollars.
Make a plan.
Selling an item for 40% off is the same as selling an item for 60% of its original price. Use this information to write and graph an equation that represents the situation.
Solve and check.
Use the plan to solve the problem. Then check your solution.
Answer:
40% = 0.40 and to find a percent of a number you multiply the number by the percent in decimal form.
So, the equation is d = 0.4p

Question 2.
Two supplementary angles have angle measures of x° and y°. Write and graph an equation that represents the relationship between the measures of the angles.
Answer:

Question 3.
A mechanic charges a diagnostic fee plus an hourly rate. The table shows the numbers of hours worked and the total costs for three customers.A fourth customer pays $285. Find the number of hours that the mechanic worked for the fourth customer.

Answer:

Performance Task

Anatomy of a Hurricane
At the beginning of this chapter, you watched a STEAM Video called “Hurricane!” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.

Graphing and Writing Linear Equations Chapter Review

Review Vocabulary
Write the definition and give an example of each vocabulary term.

Graphic Organizers
You can use a Definition and Example Chart to organize information about a concept. Here is an example of a Definition and Example Chart for the vocabulary term linear equation.

Choose and complete a graphic organizer to help you study the concept.

1. slope
2. slope of parallel lines
3. proportional relationship
4. slope-intercept form
5. standard form
6. point-slope form

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.

4.1 Graphing Linear Equations (pp. 141–146)
Learning Target: Graph linear equations.Graph the linear equation.

Question 1.
y = \(\frac{3}{5}\)x
Answer:

Question 2.
y = -2
Answer:

Question 3.
y = 9 – x
Answer:

Question 4.
y = -0.25x + 4
Answer:

Question 5.
y = \(\frac{2}{3}\)x + 2
Answer:

Question 6.
x = -5
Answer:

Question 7.
The equation y = 0.53x + 3 represents the cost y (in dollars) of riding in a taxi x miles.

a. Use a graph to estimate how much it costs to ride 5.25 miles in a taxi.
b. Use the equation to find exactly how much it costs to ride 5.25 miles in a taxi.
Answer:

y = 0.5x + 3
y = 0.5(5.25) + 3
y ≈ 5.6

Question 8.
The equation y = 9.5x represents the earnings y (in dollars) of an aquarium gift shop employee that works x hours.

a. Graph the linear equation.
b. How much does the employee earn for working 40 hours?
Answer:

Determine y for x = 40:
y = 9.5x
y = 9.5(40) = 380

Question 9.
Is y = x² a linear equation? Explain your reasoning.
Answer:
y = x²
The graph of the given equation passes through the origin, but is not linear, therefore it is not a linear equation.
So, the answer is no.

Question 10.
The sum S of the exterior angle measures of a polygon with n sides is S = 360°.
a. Plot four points (n, S) that satisfy the equation. Is the equation a linear equation? Explain your reasoning.
b. Does the value n = 2 make sense in the context of the problem? Explain your reasoning.
Answer:

The value n = 2 does not make sense in the context of the problem because a polygon has at least 3 sides.

4.2 Slope of a Line (pp. 147–154)
Learning Target: Find and interpret the slope of a line.

Describe the slope of the line. Then find the slope of the line.
Question 11.

Answer:
(x1, y1) = (3, 1)
(x2, y2) = (-3, -3)
m = (y2 – y1)/(x2 – x1)
m = (-3 – 1)/(-3 – 3)
m = -4/-6
m = 2/3

Question 12.

Answer:
(x1, y1) = (0, 4)
(x2, y2) = (2, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – 4)/(2 – 0)
m = -6/2
m = -3
The slope is negative

Find the slope of the line through the given points.
Question 13.
(-5, 4), (8, 4)
Answer:
(x1, y1) = (-5, 4)
(x2, y2) = (8, 4)
m = (y2 – y1)/(x2 – x1)
m = (4 – 4)/(8 – (-5))
m = 0/13
m = 0

Question 14.
(-3, 5), (-3, 1)
Answer:
(x1, y1) = (-3, 5)
(x2, y2) = (-3, 1)
m = (y2 – y1)/(x2 – x1)
m = (1 – 5)/(-3 + 3)
m = -4/0
m = undefined

The points in the table lie on a line. Find the slope of the line.
Question 15.

Answer:
(x1, y1) = (0, -1)
(x2, y2) = (1, 0)
m = (y2 – y1)/(x2 – x1)
m = (0 – (-1))/(1 – 0)
m = 1/1
m = 1

Question 16.

Answer:
(x1, y1) = (-2, 3)
(x2, y2) = (0, 4)
m = (y2 – y1)/(x2 – x1)
m = (4 – 3)/(0 – (-2))
m = 1/2

Question 17.
How do you know when two lines are parallel? Use an example to justify your answer.
Answer:
Two lines are parallel when their slopes are the same. In order for the two lines not to coincide, we must add the condition that their y – intercepts.
Example 1:
d1: y = 3x – 6
d2: 3x – y = 6
The lines d1 and d2 have the same slope and the same y – intercept, therefore they coincide.

Question 18.
Draw a line through the point (-1, 2) that is parallel to the graph of the line in Exercise 11.
Answer:
y = 2/3 x – 1
A (-1, 2)
y = 2/3 x + b
y = 2/3 (-1) + b
b = 8/3
The equation of d1 is:
y = 2/3 x + 8/3
Determine the x intercept of d1:
0 = 2/3 x + 8/3
0 = 2x + 8
2x = -8
x = -8/2
x = -4

4.3 Graphing Proportional Relationships (pp. 155–160)
Learning Target: Graph proportional relationships.

Tell whether x and y are in a proportional relationship. Explain your reasoning. If so, write an equation that represents the relationship.
Question 19.

Answer:
x and y are not in a proportional relationship because the line does not pass through the origin.

Question 20.

Answer:
x and y are in a proportional relationship because the line does passes through the origin.
Determine the slope k using two points from the graph
k = (10 – 0)/(2 – 0)
k = 10/2
k = 5x

Question 21.
The cost y (in dollars) to provide food for guests at a dinner party is proportional to the number x of guests attending the party. It costs $30 to provide food for 4 guests.
a. Write an equation that represents the situation.
b. Interpret the slope of the graph of the equation.
c. How much does it cost to provide food for 10 guests? Justify your answer.
Answer:
y = kx
30 = 4k
k = 30/4
k = 7.5
y = 7.5x
b. The slope 7.5 represents the unit cost for a guest.
y = 7.5 × 10
y = 75
c. Determine y for x = 10
So it costs $75 to provide food for 10 guests.

Question 22.
The distance y (in miles) you run after weeks is represented by the equation y =8x. Graph the equation and interpret the slope.
Answer:
y = 8x

Question 23.
You research that hair grows 15 centimeters per year on average. The table shows your friend’s hair growth.

a. Does your friend’s hair grow faster than average? Explain.

Answer:
The rate of growth on average is
15/12 = 1.25 cm/month
The slope/rate of growth for your friend is
(6 – 3)/(4 – 2) = 3/2 = 1.5 cm/month
As 1.5 > 1.25, your friends hair grows faster than average.

b. In the same coordinate plane, graph the average hair growth and the hair growth of your friend. Compare and interpret the steepness of each of the graphs.
Answer:
The equation for the average growth is
y = 1.25x
The equation for the friends growth is
y = 1.5 x

4.4 Graphing Linear Equations in Slope-Intercept Form (pp. 161–166)
Learning Target: Graph linear equations in slope-intercept form.

Find the slope and the -intercept of the graph of the linear equation.
Question 24.
y = -4x + 1
Answer:
y = mx + b
slope = -4 and y – intercept = 1

Question 25.
y = \(\frac{2}{3}\)x – 12
Answer:
y = mx + b
slope = \(\frac{2}{3}\) and y – intercept = -12

Question 26.
y – 7 = 0.5x
Answer:
Given the equation
y – 7 = 0.5x
y = 0.5x + 7
slope = 0.5 and y – intercept = 7

Graph the linear equation. Identify the -intercept.
Question 27.
y = 2x – 6
Answer:
Given the equation
y = 2x – 6
Comparing the above equation with slope – intercept equation
slope = 2, y – intercept = -6
Slope = rise/run = 2/1
Plot the point that is 1 unit right and 2 units up from (0, -6) = (1, -4)

The line crosses the x – axis at (3, 0)
So, the x – intercept is 3.

Question 28.
y = -4x + 8
Answer:
y = -4x + 8
slope = -4 and y – intercept = 8
So plot (0, 8)
Slope = rise/run = -4/1
plot the point that is 1 unit right and 4 units down from (0, 8) = (1, 4)

The line crosses the x- axis at (2, 0)
So the x – intercept is 2.

Question 29.
y = -x – 8
Answer:
Given the equation
y = -x – 8
comparing the above equation with sloope – intercept equation.
Slope = -1 and y – intercept = -8
Slope = rise/run = -1/1
Plot the point that is 1 unit right and 1 unit down from (0, -8) = (1, -9)

The line crosses the x-axis at (-8, 0)
So, the intercept is -8.

Question 30.
The cost y (in dollars) of one person buying admission to a fair and going on x rides is y = x + 12.

a. Graph the equation.
b. Interpret the y-intercept and the slope.
Answer:
y = x + 12
Comparing the above equation with slope – intercept equation.
Slope = 1 and y – intercept = 12
So plot (0, 20)
Slope = rise/run = 1/1
Plot the point that is 1 unit right and 1 unit up from (0, 12) = (1, 13)

The y – intercept is 12 so the initial cost of admission is $12.
The slope is 1 so for each ride the cost of the person increases $1 per ride.

Question 31.
Graph the linear equation with slope -5 and y-intercept 0.
Answer:
y – intercept = 0. So plot (0, 0)
Plot the point that is 1 unit right and 5 unit down from (0, 0) = (1, -5)

4.5 Graphing Linear Equations in Standard Form (pp. 167–172)
Learning Target: Graph linear equations in standard form.

Write the linear equation in slope-intercept form.
Question 32.
4x + 2y = -12
Answer:
4x + 2y = -12
2y = -12 – 4x
y = -6 – 2x
y = -2x – 6

Question 33.
x – y = \(\frac{1}{4}\)
Answer:
Given the equation
x – y = \(\frac{1}{4}\)
y = x – \(\frac{1}{4}\)

Graph the linear equation.
Question 34.
\(\frac{1}{4}\)x + y = 3
Answer:
\(\frac{1}{4}\)x + y = 3
y = 3 – \(\frac{1}{4}\)x
y = –\(\frac{1}{4}\)x + 3
Slope = –\(\frac{1}{4}\) and y – intercept = 3
So plot (0, 3)
Slope = rise/run = –\(\frac{1}{4}\)
Plot the point that is 4 units right and 1 unit down from (0, 3) = (4, 2)

Question 35.
-4x + 2y = 8
Answer:
-4x + 2y = 8
2y = 8 + 4x
y = 2x + 4
Slope = 2 and y – intercept = 4
So plot (0, 4)
Slope = rise/run = 2/1
Plot the point that is 1 unit right and 2 units up from (0, 4) = (1, 6)

Question 36.
x + 5y = 10
Answer:
x + 5y = 10
5y = -x + 10
y = -1/5 x + 2
Slope = -1/5 and y – intercept = 2
So, plot (0, 2)
Slope = rise/run = -1/5
Plot the point that is 5 unit right and 1 unit down from (0, 2) = (5, 1)

Question 37.
–\(\frac{1}{2}\)x + \(\frac{1}{8}\)y = \(\frac{3}{4}\)
Answer:
–\(\frac{1}{2}\)x + \(\frac{1}{8}\)y = \(\frac{3}{4}\)
\(\frac{1}{8}\)y = \(\frac{3}{4}\)  + \(\frac{1}{2}\)x
y = 4x + 6
Slope = 4 and y – intercept = 6
So plot (0, 6)
Slope = rise/run = 4/1
Plot the point that is 1 unit right and 4 units up from (0, 6) = (1, 10)

Question 38.
A dog kennel charges $30 per night to board your dog and $6 for each hour of playtime. The amount of money you spend is given by 30x + 6y = 180, where x is the number of nights and y is the number of hours of playtime. Graph the equation and interpret the intercepts.
Answer:
Given,
A dog kennel charges $30 per night to board your dog and $6 for each hour of playtime.
The amount of money you spend is given by 30x + 6y = 180,
where x is the number of nights and y is the number of hours of playtime.
30x + 6y = 180
6y = -30x + 180
y = -5x + 30

The x – intercept is 6, which means that the dog can stay for 6 nights when there is no playtime.
The y – intercept is 30, which means the dog can play for 30 hours when he does not spend any night at the kennel.

4.6 Writing Equations in Slope-Intercept Form (pp. 173–178)
Learning Target: Write equations of lines in slope-intercept form.

Write an equation in slope-intercept form of the line that passes through the given points.
Question 39.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (1 – (-2))/(3 – 0)
m = (1 + 2)/(3 – 0)
m = 3/3
m = 1
We have to find the y – intercept because the line crosses the y – axis at (0, -2)
y = mx + b
y = x – 2

Question 40.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (4 – 2)/(0 – 4)
m = 2/-4
m = -1/2
We have to find the y – intercept because the line crosses the y – axis at (0, 4)
y = mx + b
y =-1/2 x + 4

Question 41.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (-2 – 1)/(2 – 0)
m = -3/2
We have to find the y – intercept because y = 1 when x = 0, the y – intercept is 1.
y = mx + b
y = -3/2 x + 1

Question 42.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (-1 – (-3))/(1 – 0)
m = 2/1
m = 2
We have to find the y – intercept because y = -3 when x = 0, the y – intercept is -3.
y = mx + b
y = 2x + (-3)
y = 2x – 3

Question 43.
Write an equation of the line that passes through (0, 8) and (6, 8).
Answer:
m = (y2 – y1)/(x2 – x1)
m = (8 – 8)/(6 – 0)
m = 0/6
m = 0
We have to find the y – intercept because y = 8 when x = 0, the y – intercept is 8.
y = mx + b
y = (0) x + 8
y = 8

Question 44.
Write an equation of the line that passes through (0, -5) and (-5, -5).
Answer:
m = (y2 – y1)/(x2 – x1)
m = (-5 – (-5))/(-5 – 0)
m = 0/-5
m = 0
We have to find the y – intercept because y = -5 when x = 0, the y – intercept is -5
y = mx + b
y = (0) x + (-5)
y = -5

Question 45.
A construction crew is extending a highway sound barrier that is 13 miles long. The crew builds \(\frac{1}{2}\) of a mile per week. Write an equation in slope -intercept form that represents the length y (in miles) of the barrier after x weeks.
Answer:
Given,
A construction crew is extending a highway sound barrier that is 13 miles long.
The crew builds \(\frac{1}{2}\) of a mile per week.
y = mx + b
m = \(\frac{1}{2}\)
b = 13
y = \(\frac{1}{2}\)x + 13

4.7 Writing Equations in Point-Slope Form (pp. 179–184)
Learning Target: Write equations of lines in point-slope form.

Write an equation in point-slope form of the line that passes through the given point and has the given slope.
Question 46.
(4, 4); m = 3
Answer:
y – y1 = m(x – x1)
Substitute m value, x and y value in the equation
y – (4) = 3(x – 4)
y – 4 = 3(x – 4)

Question 47.
(2, -8); m = –\(\frac{2}{3}\)
Answer:
y – y1 = m(x – x1)
Substitute m value, x and y value in the equation
y – (-8) = –\(\frac{2}{3}\)(x – 2)
y + 8 = –\(\frac{2}{3}\)(x – 2)

Write an equation in slope-intercept form of the line that passes through the given points.
Question 48.
(-4, 2), (6, -3)
Answer:
m = (y2 – y1)/(x2 – x1)
m = (-3 – 2)/(6 – (-4))
m = -5/10
m = -1/2
y – y1 = m(x – x1)
Substitute m value, x and y value in the equation
y – 2 = –\(\frac{1}{2}\)(x – (-4))
y – 2 = –\(\frac{1}{2}\)(x + 4)
y = –\(\frac{1}{2}\)x

Question 49.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (5 – 1)/(3 – 2)
m = 4/1
m = 4
y – y1 = m(x – x1)
Substitute m value, x and y value in the equation
y – (-3) = 4(x – 1)
y + 3 = 4x – 4
y = 4x – 7

Question 50.
The table shows your elevation y (in feet) on a ski slope after x minutes.

a. Write an equation that represents your elevation after x minutes.

Answer:
m = (600 – 800)/(2 – 1)
m = -200
800 = -200(1) + b
800 = -200 + b
800 + 200 = b
b = 1000 feet

b. What is your starting elevation?

Answer:
The starting elevation is the y – intercept
b = 1000 feet

c. After how many minutes do you reach the bottom of the ski slope? Justify your answer.
Answer:
0 = -200x + 1000
0 – 1000 = -200x
-1000 = -200x
200x = 1000
x = 5 minutes

Question 51.
A company offers cable television at$29.95 per month plus a one-time installation fee. The total cost for the first six months of service is $214.70. a. Write an equation in point-slope form that represents the total cost you pay for cable television after x months.
b. How much is the installation fee? Justify your answer.
Answer:
y – y1 = m(x – x1)
m = 29.95
y – 214.70 = 29.95(x – 6)
y – 214.70 + 214.70 = 29.95x – 179.97 + 2147.70
y = 29.95x + 35
b = 35

Question 52.
When might it be better to represent an equation in point-slope form rather than slope-intercept form? Use an example to justify your answer.
Answer:
When we are given the slope and a point that is the y – intercept, then the easiest way is to use the slope – intercept form y = mx + b
Example:
m = 2
(0, 5)
y = 2x + 5
m = 2
(1, 3)
y – 3 = 2(x – 1)
Easier when given the slope and a point that is not the y – intercept.

Graphing and Writing Linear Equations Practice Test

Find the slope and the -intercept of the graph of the linear equation.
Question 1.
y = 6x – 5
Answer:
y = 6x – 5
Slope = 6 and y – intercept = -5

Question 2.
y – 1 = 3x + 8.4
Answer:
Given the equation
y – 1 = 3x + 8.4
y = 3x + 8.4 + 1
y = 3x + 9.4
Slope = 3 and y – intercept = 9.4

Question 3.
–\(\frac{1}{2}\)x + 2y = 7
Answer:
Given the equation
–\(\frac{1}{2}\)x + 2y = 7
y = \(\frac{1}{4}\)x + \(\frac{7}{2}\)
Slope = \(\frac{1}{4}\) and y – intercept = \(\frac{7}{2}\)

Graph the linear equation.
Question 4.
y = –\(\frac{1}{2}\)x – 5
Answer:
Given the equation
y = –\(\frac{1}{2}\)x – 5
Slope = –\(\frac{1}{2}\) and y – intercept = -5
So plot (0, -5)
Plot the point that is 2 units right and 1 unit down from (0, -5) = (2, -6)
Draw a line through the two points.

Question 5.
-3x + 6y = 12
Answer:
Given the equation
-3x + 6y = 12
6y = 3x + 12
y = \(\frac{1}{2}\)x + 2
Slope = \(\frac{1}{2}\), y – intercept = 2
Slope = rise/run = \(\frac{1}{2}\)
Plot the point that is 2 units right and 1 unit up from (0, 2) = (2, 3)

Question 6.
y = \(\frac{2}{3}\)x
Answer:
Given the equation
y = \(\frac{2}{3}\)x
Slope = \(\frac{2}{3}\), y – intercept = 0
Slope = rise/run = \(\frac{2}{3}\)
Plot the point that is 3 units right and 2 unit up from (0, 0) = (3, 2)

Question 7.
Which lines are parallel? Explain.

Answer:
Red line:
(x1, y1) = (-4, 1)
(x2, y2) = (2, 4)
m = (y2 – y1)/(x2 – x1)
m = (4 – 1)/(2 – (-4))
m = 3/6
m = 1/2
Blue line:
(x1, y1) = (-4, -1)
(x2, y2) = (2, 0.5)
m = (y2 – y1)/(x2 – x1)
m = (0.5 – (-1))/(2 – (-4))
m = 1.5/6
m = 1/4
Green Line:
(x1, y1) = (-2, -4)
(x2, y2) = (2, -2)
m = (y2 – y1)/(x2 – x1)
m = (-2 – (-4))/(2 – (-2))
m = 2/4
m = 1/2
Red lines and Green lines are parallel because both have same slope = 1/2

Question 8.
The points in the table lie on a line. Find the slope of the line.

Answer:
(x1, y1) = (-1, -4)
(x2, y2) = (0, -1)
m = (y2 – y1)/(x2 – x1)
m = (-1 – (-4))/(0 – (-1))
m = 3/1
m = 3

Write an equation in slope-intercept form of the line that passes through the given points.
Question 9.

Answer:
(x1, y1) = (-4, 1)
(x2, y2) = (2, 4)
m = (y2 – y1)/(x2 – x1)
m = (-5 – (-1))/(3 – 0)
m = -4/3
Because the line crosses the y – axis at (0, -1), the y – intercept is -1.
y = mx + b
y = -4/3x – 1

Question 10.

Answer:
m = (y2 – y1)/(x2 – x1)
m = (2 – 2)/(0 – (-2))
m = 0/2
m = 0
Because y = 2 when x =0, the y – intercept is 2.
y = mx + b
y = 2

Question 11.
Write an equation in point-slope form of the line that passes through (-4, 1) and (4, 3).
Answer:
m = (y2 – y1)/(x2 – x1)
m = (3 – 1)/(4 – (-4))
m = 2/8
m = 1/4
y – y1 = m(x – x1)
y – 1 = 1/4(x – (-4))
y – 1 = 1/4(x + 4)

Question 12.
The number y of new vocabulary words that you learn after x weeks is represented by the equation y = 15x.
a. Graph the equation and interpret the slope.
b. How many new vocabulary words do you learn after 5 weeks?
c. How many more vocabulary words do you learn after 6 weeks than after 4 weeks?
Answer:
a.
b. y = 15 . 5
y = 75 words
c. 15 . 6 – 15 . 4 = 90 – 60 = 30 words

Question 13.
You used $90 worth of paint for a school float. The amount of money you spend is given by 18x + 15y = 90, where x is the number of gallons of blue paint and y is the number of gallons of white paint. Graph the equation and interpret the intercepts.
Answer:
Given,
18x + 15y = 90
15y = -18x + 90
y = -6/5 x + 6

The x – intercept is 5 and shows that 5 gallons of blue paint might be bought when no gallon of the white pants is bought.
The y – intercept is 6 and shows that 6 gallons of white paint might be bought when no gallon of blue is bought.

Graphing and Writing Linear Equations Cumulative Practice

Question 1.
Which equation matches the line shown in the graph?

A. y =2x – 2
B. y = 2x + 1
C. y = x – 2
D. y = x + 1
Answer:
m = (-2 – 0)/(0 – 1)
m = 2
y = 2x – 2
Thus the correct answer is option A.

Question 2.
Which point lies on the graph of 6x – 5y = 14?
F. (-4, -1)
G. (-2, 4)
H. (-1, -4)
I. (4, -2)
Answer:
6x – 5y = 14
F. 6(-4) – 5(-1) = 14
-24 + 5 = 14
-19 ≠ 14
G. 6(-2) – 5(4) = 14
-12 – 20 = 14
-32 ≠ 14
H. 6(-1) – 5(-4) = 14
-6 + 20 = 14
14 = 14
Thus the correct answer is option H.

Question 3.
You reflect the triangle in the x-axis. What are the coordinates of the image?

A. X'(4, 1), Y'(2, 3), Z'(-2, 1)
B. X'(4, -1), Y'(2, -3), Z'(-2, -1)
C. X'(-4, -1), Y(-2, -3), Z'(2, -1)
D. X'(1, 4), Y'(3, 2), Z'(1, -2)
Answer:

Thus the correct answer is option C.

Question 4.
Which of the following is the equation of a line parallel to the line shown in the graph?

Answer:
m = (-4 – 2)/(6 – 4)
m = -6/2
m = -3
Two lines parallel if they have the same slope.
From the given equations, the one having the slope -3 is y = -3x + 5
Thus the correct answer is option H.

Question 5.
What is the value of x?

Answer:
122 = 47 + x
47 + x = 122
x = 122 – 47
x = 75

Question 6.
An emergency plumber charges $49.00 plus $70.00 per hour of the repair. A bill to repair your sink is $241.50. This can be modeled by 70.00 h + 49.00 = 241.50, where h represents the number of hours for the repair. How many hours did it take to repair your sink?
A. 2.75 hours
B. 3.45 hours
C. 4.15 hours
D. 13,475 hours
Answer:
70.00 h + 49.00 = 241.50
70h = 241.50 – 49
70h = 192.5
h = 2.75 hours
Thus the correct answer is option A.

Question 7.
It costs $40 to rent a car for one day. In addition, the rental agency charges you for each mile driven, as shown in the graph.

Part A Determine the slope of the line joining the points on the graph.
Part B Explain what the slope represents.
Answer:
m = (50 – 40)/(100 – 0)
m = 10/100
m = 0.1

Question 8.
What value of makes the equation true?

7 + 2x = 4x – 5
Answer:
7 + 2x = 4x – 5
2x – 4x = -5 – 7
-2x = -12
x = 6

Question 9.
Trapezoid KLMN is graphed in the coordinate plane shown.

Rotate Trapezoid KLMN 90° clockwise about the origin. What are the coordinates of point M’, the image of point M after the rotation?
F. (-3, -2)
G. (-2, -3)
H. (-2, 3)
I. (3, 2)
Answer: M'(-3, -2)
Thus the correct answer is option F.

Question 10.
Solve the formula K = 3M – 7.
A. M = K + 7
B. M = \(\frac{K+7}{3}\)
C. M = \(\frac{K}{3}\) + 7
D. M = \(\frac{K-7}{3}\)
Answer:
K = 3M – 7
K + 7 = 3M
M = \(\frac{K+7}{3}\)
Thus the correct answer is option B.

Question 11.
What is the distance across the canyon?

F. 3.6 ft
G. 12 ft
H. 40 ft
I. 250 ft
Answer:
100/30 = d/12
3d = 12 × 10
3d = 120
d = 40 feet
Thus the correct answer is option H.

Conclusion:

All the solutions in the above article are beneficial for all the students of middle school students. All the solutions are prepared by the math professionals. The solutions are given clearly with step by step explanations. If you have any doubts regarding the chapter we are always ready to clarify your doubts. All you have to do is to post the comments in the below comment box.