# Points A and B are at (6 ,5 ) and (3 ,8 ), respectively. Point A is rotated counterclockwise about the origin by pi/2  and dilated about point C by a factor of 2 . If point A is now at point B, what are the coordinates of point C?

Jun 8, 2018

color(green)("Coordinates of " C = (-13,4)

#### Explanation:

$A \left(6 , 5\right) , B \left(3 , 8\right) , \text{ rotated counter clockwise by " pi/2 " and dilated by factor 2}$

Coordinates of A after $\frac{\pi}{2}$ counter clockwise rotation is

$A \left(6 , 5\right) \to A ' \left(- 5 , 6\right)$

$\vec{B C} = 2 \vec{A ' C}$

$b - c = 2 \left(a ' - c\right)$

$c = 2 a ' - b$

C((x),(y)) = 2 ((-5),(6)) - ((3),(8)) =color(green)( ((-13),(4))

Jun 8, 2018

$C = \left(- 13 , 4\right)$

#### Explanation:

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

• " a point "(x,y)to(-y,x)

$A \left(6 , 5\right) \to A ' \left(- 5 , 6\right) \text{ where A' is the image if A}$

$\vec{C B} = \textcolor{red}{2} \vec{C A '}$

$\underline{b} - \underline{c} = 2 \left(\underline{a} ' - \underline{c}\right)$

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

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

$\textcolor{w h i t e}{\underline{c}} = 2 \left(\begin{matrix}- 5 \\ 6\end{matrix}\right) - \left(\begin{matrix}3 \\ 8\end{matrix}\right)$

$\textcolor{w h i t e}{\underline{c}} = \left(\begin{matrix}- 10 \\ 12\end{matrix}\right) - \left(\begin{matrix}3 \\ 8\end{matrix}\right) = \left(\begin{matrix}- 13 \\ 4\end{matrix}\right)$

$\Rightarrow C = \left(- 13 , 4\right)$