Points A and B are at (6 ,3 ) and (1 ,9 ), respectively. Point A is rotated counterclockwise about the origin by pi  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 16, 2018

$C = \left(- \frac{19}{2} , - 9\right)$

Explanation:

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

$\text{a point } \left(x , y\right) \to \left(- x , - y\right)$

$\Rightarrow A \left(6 , 3\right) \to A ' \left(- 6 , - 3\right) \text{ where A' is the image of A}$

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

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

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

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

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = 3 \left(\begin{matrix}- 6 \\ - 3\end{matrix}\right) - \left(\begin{matrix}1 \\ 9\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow 2 \underline{c}} = \left(\begin{matrix}- 18 \\ - 9\end{matrix}\right) - \left(\begin{matrix}1 \\ 9\end{matrix}\right) = \left(\begin{matrix}- 19 \\ - 18\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{2} \left(\begin{matrix}- 19 \\ - 18\end{matrix}\right) = \left(\begin{matrix}- \frac{19}{2} \\ - 9\end{matrix}\right)$

$\Rightarrow C = \left(- \frac{19}{2} , - 9\right)$