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# Parallelogram ABCD has vertices$A \left( - 2 , - 5 \frac { 1 } { 2 } \right) , B \left( - 4 , - 5 \frac { 1 } { 2 } \right)$, C(-3, -2), and D(-1, -2). Find the vertices of parallelogram$A ^ { \prime } B ^ { \prime } C D ^ { \prime }$after a transformation of$2 \frac { 1 } { 2 }$units down.

Solution

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The transformation is a translation of $2\dfrac{1}{2}$ down. This means that $y$ coordinate change.

Subtract $2\dfrac{1}{2}$ from the $y$ coordinate.

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

A'$(-2, -8)$

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

B'$(-4, -8)$

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

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

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

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

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