Solved unit 9 transformations homework 7 sequences of | Chegg.com - Free Printable

Let’s solve each problem step by step. We’ll do the transformations one at a time and track where each point ends up.

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Problem 1: Rectangle DEFG with vertices D(-2,7), E(2,3), F(0,1), G(-4,5)

We are doing two transformations in order:

a) Translation along the rule (x, y) → (x + 6, y - 8)
b) Reflection in the y-axis

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Start with original points:

D(-2, 7)
E(2, 3)
F(0, 1)
G(-4, 5)

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Step a: Apply translation (x+6, y-8)

Add 6 to x, subtract 8 from y for each point.

→ D’ after translation: (-2+6, 7-8) = (4, -1)
→ E’ after translation: (2+6, 3-8) = (8, -5)
→ F’ after translation: (0+6, 1-8) = (6, -7)
→ G’ after translation: (-4+6, 5-8) = (2, -3)

Now we have intermediate points:
D’(4,-1), E’(8,-5), F’(6,-7), G’(2,-3)

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Step b: Reflect over the y-axis

Reflection over y-axis means: (x, y) → (-x, y)

So flip the sign of the x-coordinate.

→ D'' = (-4, -1)
→ E'' = (-8, -5)
→ F'' = (-6, -7)
→ G'' = (-2, -3)

These are the final image coordinates.

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Final Answer for Problem 1:

D''(-4, -1)
E''(-8, -5)
F''(-6, -7)
G''(-2, -3)

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Problem 2: Triangle LMN with vertices L(6,6), M(8,8), N(8,3)

Transformations:

a) Reflection in the line x = 5
b) 270° counterclockwise rotation about the origin

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Original points:

L(6,6)
M(8,8)
N(8,3)

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Step a: Reflect over the line x = 5

To reflect a point over vertical line x = 5:

Find how far the point is from x=5, then go that same distance on the other side.

Formula: If reflecting (x,y) over x = a, new x = 2a - x, y stays same.

Here, a = 5 → new x = 10 - x

→ L’ after reflection: (10 - 6, 6) = (4, 6)
→ M’ after reflection: (10 - 8, 8) = (2, 8)
→ N’ after reflection: (10 - 8, 3) = (2, 3)

Intermediate points:
L’(4,6), M’(2,8), N’(2,3)

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Step b: Rotate 270° counterclockwise about origin

Rule for 270° CCW rotation: (x, y) → (y, -x)

Apply to each:

→ L'' = (6, -4)
→ M'' = (8, -2)
→ N'' = (3, -2)

Check:

For L’(4,6): (x=4, y=6) → (6, -4) ✔️
For M’(2,8): (2,8) → (8, -2) ✔️
For N’(2,3): (2,3) → (3, -2) ✔️

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Final Answer for Problem 2:

L''(6, -4)
M''(8, -2)
N''(3, -2)

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Problem 3: Quadrilateral ABCD with vertices A(0,6), B(-3,-6), C(-9,-6), D(-12,-3)

Transformations:

a) Dilation with scale factor 1/3 centered at origin
b) Translation along vector <-5, -1>

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Original points:

A(0,6)
B(-3,-6)
C(-9,-6)
D(-12,-3)

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Step a: Dilate by 1/3 from origin

Multiply both x and y by 1/3.

→ A’ = (0 * 1/3, 6 * 1/3) = (0, 2)
→ B’ = (-3 * 1/3, -6 * 1/3) = (-1, -2)
→ C’ = (-9 * 1/3, -6 * 1/3) = (-3, -2)
→ D’ = (-12 * 1/3, -3 * 1/3) = (-4, -1)

Intermediate points:
A’(0,2), B’(-1,-2), C’(-3,-2), D’(-4,-1)

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Step b: Translate by vector <-5, -1>

That means: add -5 to x, add -1 to y → (x - 5, y - 1)

→ A'' = (0 - 5, 2 - 1) = (-5, 1)
→ B'' = (-1 - 5, -2 - 1) = (-6, -3)
→ C'' = (-3 - 5, -2 - 1) = (-8, -3)
→ D'' = (-4 - 5, -1 - 1) = (-9, -2)

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Final Answer for Problem 3:

A''(-5, 1)
B''(-6, -3)
C''(-8, -3)
D''(-9, -2)

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Final Answer:
1. D''(-4, -1), E''(-8, -5), F''(-6, -7), G''(-2, -3)
2. L''(6, -4), M''(8, -2), N''(3, -2)
3. A''(-5, 1), B''(-6, -3), C''(-8, -3), D''(-9, -2)