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, carefully tracking how each point moves.

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Problem 4: Triangle XYZ with vertices X(-6, 4), Y(0, 6), Z(-4, 2)

We are doing two transformations in order:
a) 180° counterclockwise rotation about the origin
b) dilation with scale factor 3/2 centered at the origin

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Step 1: 180° rotation about origin

Rule for 180° rotation (either direction — it’s the same):
→ (x, y) → (-x, -y)

Apply to each point:

- X(-6, 4) → X’ = (6, -4)
- Y(0, 6) → Y’ = (0, -6)
- Z(-4, 2) → Z’ = (4, -2)

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Step 2: Dilation by scale factor 3/2 from origin

Rule: Multiply both x and y coordinates by 3/2.

Apply to rotated points:

- X’(6, -4) → X'' = (6 × 3/2, -4 × 3/2) = (9, -6)
- Y’(0, -6) → Y'' = (0 × 3/2, -6 × 3/2) = (0, -9)
- Z’(4, -2) → Z'' = (4 × 3/2, -2 × 3/2) = (6, -3)

So final answers for Problem 4:

X''(9, -6)
Y''(0, -9)
Z''(6, -3)

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Problem 5: Parallelogram PQRS with vertices P(-3, -4), Q(0, -3), R(1, -8), S(-4, -9)

Transformations:
a) translation along vector <7, 0>
b) 90° counterclockwise rotation about (-3, -2)

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Step 1: Translation by <7, 0>

Add 7 to x-coordinate, leave y unchanged.

- P(-3, -4) → P’ = (-3+7, -4) = (4, -4)
- Q(0, -3) → Q’ = (0+7, -3) = (7, -3)
- R(1, -8) → R’ = (1+7, -8) = (8, -8)
- S(-4, -9) → S’ = (-4+7, -9) = (3, -9)

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Step 2: 90° CCW rotation about point (-3, -2)

This is trickier because the center is NOT the origin.

General method for rotating around any point (h,k):

1. Translate so that (h,k) becomes origin: subtract h from x, k from y.
2. Apply 90° CCW rotation rule: (x,y) → (-y, x)
3. Translate back: add h to x, k to y.

Center: (h,k) = (-3, -2)

Rotation formula combined:

For any point (x,y), after 90° CCW about (-3,-2):

New x = - (y - (-2)) + (-3) = -(y + 2) - 3
New y = (x - (-3)) + (-2) = (x + 3) - 2

Simplify:

New x = -y - 2 - 3 = -y - 5
New y = x + 3 - 2 = x + 1

Wait — let me double-check this derivation.

Actually, better to do step-by-step per point to avoid mistakes.

Take point P’(4, -4)

Step A: Shift so center (-3,-2) is origin:
P’_shifted = (4 - (-3), -4 - (-2)) = (7, -2)

Step B: Rotate 90° CCW: (x,y) → (-y, x)
→ (-(-2), 7) = (2, 7)

Step C: Shift back: add (-3, -2)
→ (2 + (-3), 7 + (-2)) = (-1, 5)

So P'' = (-1, 5)

Now Q’(7, -3)

Shift: (7 - (-3), -3 - (-2)) = (10, -1)
Rotate: (-(-1), 10) = (1, 10)
Shift back: (1 + (-3), 10 + (-2)) = (-2, 8)

Q'' = (-2, 8)

R’(8, -8)

Shift: (8 - (-3), -8 - (-2)) = (11, -6)
Rotate: (-(-6), 11) = (6, 11)
Shift back: (6 + (-3), 11 + (-2)) = (3, 9)

R'' = (3, 9)

S’(3, -9)

Shift: (3 - (-3), -9 - (-2)) = (6, -7)
Rotate: (-(-7), 6) = (7, 6)
Shift back: (7 + (-3), 6 + (-2)) = (4, 4)

S'' = (4, 4)

Double-checking one more time for P’:

Original after translate: (4, -4)
Center: (-3, -2)
Vector from center to P’: (4 - (-3), -4 - (-2)) = (7, -2)
Rotate 90° CCW: becomes (2, 7) [since (a,b) → (-b,a)]
Then add back center: (-3 + 2, -2 + 7) = (-1, 5) ✔

All good.

Final answers for Problem 5:

P''(-1, 5)
Q''(-2, 8)
R''(3, 9)
S''(4, 4)

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Problem 6: Triangle CDE with vertices C(-2, -2), D(1, -4), E(-3, -4)

Transformations:
a) dilation with scale factor 4 centered at (-1, -3)
b) reflection in the line y = -x

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Step 1: Dilation by 4 from center (-1, -3)

Again, not centered at origin. Use similar method as before.

For any point (x,y), dilate by 4 from (h,k)=(-1,-3):

1. Shift: (x - h, y - k) = (x +1, y +3)
2. Scale by 4: (4(x+1), 4(y+3))
3. Shift back: (4(x+1) + h, 4(y+3) + k) = (4(x+1) -1, 4(y+3) -3)

Simplify:

New x = 4x + 4 - 1 = 4x + 3
New y = 4y + 12 - 3 = 4y + 9

Wait — let’s verify with actual steps per point to be safe.

Start with C(-2, -2)

Shift relative to center (-1,-3):
C_shifted = (-2 - (-1), -2 - (-3)) = (-1, 1)

Scale by 4: (-4, 4)

Shift back: (-4 + (-1), 4 + (-3)) = (-5, 1)

So C’ = (-5, 1)

D(1, -4)

Shift: (1 - (-1), -4 - (-3)) = (2, -1)
Scale: (8, -4)
Shift back: (8 + (-1), -4 + (-3)) = (7, -7)

D’ = (7, -7)

E(-3, -4)

Shift: (-3 - (-1), -4 - (-3)) = (-2, -1)
Scale: (-8, -4)
Shift back: (-8 + (-1), -4 + (-3)) = (-9, -7)

E’ = (-9, -7)

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Step 2: Reflection over line y = -x

Rule for reflecting over y = -x:
→ (x, y) → (-y, -x)

Apply to diluted points:

C’(-5, 1) → C'' = (-1, 5)
D’(7, -7) → D'' = (7, -7) → wait: (-(-7), -7) = (7, -7)? No!

Wait: Rule is (x,y) → (-y, -x)

So:

C’(-5, 1) → (-1, 5) ✔
D’(7, -7) → (-(-7), -7) = (7, -7)? Wait no:

(x,y) = (7, -7) → (-y, -x) = (-(-7), -7) = (7, -7) — that can’t be right? Let's think.

Actually: if you reflect (7, -7) over y = -x, what happens?

The line y = -x swaps and negates. So yes: (a,b) → (-b, -a)

So (7, -7) → (-(-7), -7) = (7, -7) — oh! It lies ON the line y = -x? Because -7 = - (7). Yes! So it stays put.

Similarly, E’(-9, -7) → (-(-7), -(-9)) = (7, 9)

Check:

C’(-5,1) → (-1, 5)
D’(7,-7) → (7, -7) [on the line]
E’(-9,-7) → (7, 9)

Wait — let me recalculate E’:

E’ was (-9, -7)

Reflect over y = -x: (x,y) → (-y, -x)

So: -y = -(-7) = 7
-x = -(-9) = 9
→ (7, 9) ✔

But D’(7, -7):
-y = -(-7) = 7
-x = -(7) = -7
→ (7, -7) — correct, since it’s on the line.

C’(-5,1):
-y = -1
-x = -(-5) = 5
→ (-1, 5) ✔

So final answers for Problem 6:

C''(-1, 5)
D''(7, -7)
E''(7, 9)

Wait — hold on! For D’, we had (7, -7), which reflects to (7, -7). But let me confirm geometry.

Point (7, -7): plug into y = -x → -7 = -7 → yes, it’s on the line, so reflection leaves it unchanged. Correct.

All set.

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Final Answer:

Problem 4:
X''(9, -6)
Y''(0, -9)
Z''(6, -3)

Problem 5:
P''(-1, 5)
Q''(-2, 8)
R''(3, 9)
S''(4, 4)

Problem 6:
C''(-1, 5)
D''(7, -7)
E''(7, 9)