# Points A and B are at (4 ,5 ) and (2 ,0 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/2  and dilated about point C by a factor of 4 . If point A is now at point B, what are the coordinates of point C?

##### 1 Answer
Jun 1, 2018

$C = \left(6 , - \frac{16}{3}\right)$

#### Explanation:

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

• " a point "(x,y)to(y,-x)

$A \left(4 , 5\right) \to A ' \left(5 , - 4\right) \text{ where A' is the image of A}$

$\vec{C B} = \textcolor{red}{4} \vec{C A '}$

$\underline{b} - \underline{c} = 4 \left(\underline{a} ' - \underline{c}\right)$

$\underline{b} - \underline{c} = 4 \underline{a} ' - 4 \underline{c}$

$3 \underline{c} = 4 \underline{a} ' - \underline{b}$

$\textcolor{w h i t e}{3 \underline{c}} = 4 \left(\begin{matrix}5 \\ - 4\end{matrix}\right) - \left(\begin{matrix}2 \\ 0\end{matrix}\right)$

$\textcolor{w h i t e}{3 \underline{c}} = \left(\begin{matrix}20 \\ - 16\end{matrix}\right) - \left(\begin{matrix}2 \\ 0\end{matrix}\right) = \left(\begin{matrix}18 \\ - 16\end{matrix}\right)$

$\underline{c} = \frac{1}{3} \left(\begin{matrix}18 \\ - 16\end{matrix}\right) = \left(\begin{matrix}6 \\ - \frac{16}{3}\end{matrix}\right)$

$\Rightarrow C = \left(6 , - \frac{16}{3}\right)$