Points A and B are at (3 ,7 ) and (6 ,1 ), respectively. Point A is rotated counterclockwise about the origin by pi/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?

Aug 15, 2017

$C \left(- \frac{27}{2} , 4\right)$

Explanation:

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

• " a point "(x,y)to(-y,x)

$\Rightarrow a \left(3 , 7\right) \to A ' \left(- 7 , 3\right) \text{ where A' is the image of A}$

$\text{under a dilatation about C of factor 3}$

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

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

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

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

$\textcolor{w h i t e}{2 \underline{c} \times x} = 3 \left(\begin{matrix}- 7 \\ 3\end{matrix}\right) - \left(\begin{matrix}6 \\ 1\end{matrix}\right)$

$\textcolor{w h i t e}{\times \times x} = \left(\begin{matrix}- 21 \\ 9\end{matrix}\right) - \left(\begin{matrix}6 \\ 1\end{matrix}\right) = \left(\begin{matrix}- 27 \\ 8\end{matrix}\right)$

$\Rightarrow \underline{c} = \frac{1}{2} \left(\begin{matrix}- 27 \\ 8\end{matrix}\right) = \left(\begin{matrix}- \frac{27}{2} \\ 4\end{matrix}\right)$

$\Rightarrow C = \left(- \frac{27}{2} , 4\right)$