# Points A and B are at (8 ,5 ) and (2 ,2 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/2  and dilated about point C by a factor of 3 . If point A is now at point B, what are the coordinates of point C?

Feb 27, 2018

$C = \left(\frac{13}{2} , - 13\right)$

#### Explanation:

$\text{under a counterclockwise rotation about the origin of } \frac{3 \pi}{2}$

• " a point "(x,y)to(y,-x)

$\Rightarrow A \left(8 , 5\right) \to A ' \left(5 , - 8\right) \text{ where A' is the image of A}$

$\Rightarrow \vec{C B} = \textcolor{red}{3} \vec{C A '}$

$\Rightarrow \underline{b} - \underline{c} = 3 \left(\underline{a} ' - \underline{c}\right)$

$\Rightarrow \underline{b} - \underline{c} = 3 \underline{a} ' - 3 \underline{c}$

$\Rightarrow 2 \underline{c} = 3 \underline{a} ' - \underline{b}$

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = 3 \left(\begin{matrix}5 \\ - 8\end{matrix}\right) - \left(\begin{matrix}2 \\ 2\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = \left(\begin{matrix}15 \\ - 24\end{matrix}\right) - \left(\begin{matrix}2 \\ 2\end{matrix}\right) = \left(\begin{matrix}13 \\ - 26\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{2} \left(\begin{matrix}- 13 \\ - 26\end{matrix}\right) = \left(\begin{matrix}\frac{13}{2} \\ - 13\end{matrix}\right)$

$\Rightarrow C = \left(\frac{13}{2} , - 13\right)$