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Question

Parallelogram ABCD has vertices

A(2,512),B(4,512)A \left( - 2 , - 5 \frac { 1 } { 2 } \right) , B \left( - 4 , - 5 \frac { 1 } { 2 } \right)

, C(-3, -2), and D(-1, -2). Find the vertices of parallelogram

ABCDA ^ { \prime } B ^ { \prime } C D ^ { \prime }

after a transformation of

2122 \frac { 1 } { 2 }

units down.

Solution

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Answered 2 years ago
Answered 2 years ago
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The transformation is a translation of 2122\dfrac{1}{2} down. This means that yy coordinate change.

Subtract 2122\dfrac{1}{2} from the yy coordinate.

A(2,512)\left(-2, 5\dfrac{1}{2}\right) \rightarrow A'(2,512212)\left(-2, -5\dfrac{1}{2}-2\dfrac{1}{2}\right)

A'(2,8)(-2, -8)

B(4,512)\left(-4, 5\dfrac{1}{2}\right) \rightarrow B'(4,512212)\left(-4, -5\dfrac{1}{2}-2\dfrac{1}{2}\right)

B'(4,8)(-4, -8)

C(3,2)\left(-3, -2\right) \rightarrow C'(3,2212)\left(-3, -2-2\dfrac{1}{2}\right)

C'(3,412)\left(-3, -4\dfrac{1}{2}\right)

D(1,2)\left(-1, -2\right) \rightarrow D'(1,2212)\left(-1, -2-2\dfrac{1}{2}\right)

D'(1,412)\left(-1, -4\dfrac{1}{2}\right)

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