# Points A and B are at (7 ,9 ) and (6 ,2 ), respectively. Point A is rotated counterclockwise about the origin by (3pi)/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?

Apr 29, 2018

The image of $A$ rotated by $\frac{3 \pi}{2}$ is $D \left(9 , - 7\right) .$ We get $B = r \left(D - C\right) + C$ and we solve giving $C = \left(12 , - 16\right) .$

#### Explanation:

I don't get why these come up as recently asked when they're two years old.

$\frac{3 \pi}{2} = {270}^{\circ}$

The image of rotating $A \left(7 , 9\right)$ by ${270}^{\circ}$ around the origin lands in the fourth quadrant, so must be $D = \left(9 , - 7\right)$.

Let's see where a point D ends up after dilation around C by a factor of $r$. One way to think about it is to translate the dilation point to the origin, scale the translate D using $r$, and translate back:

$D ' = r \left(D - C\right) + C = \left(1 - r\right) C + r D$

That is always interesting to me. It's the parametric equation for a line between C ($r = 0$) and D ($r = 1$). D' is on that line, $r$ times the distance from C than D is. That makes sense given the meaning of dilation.

We have $D ' = B$ which gives us an equation to solve for C:

$B = \left(1 - r\right) C + r D$

$B - r D = \left(1 - r\right) C$

$C = \frac{B - r D}{1 - r}$

$C = \frac{\left(6 , 2\right) - 2 \left(9 , - 7\right)}{1 - 2} = \left(12 , - 16\right)$

Check:

(1 - r) C + r D = (-12,16) + (18,-14)=(6,2)=B quad sqrt

Apr 29, 2018

$C = \left(12 , - 16\right)$

#### Explanation:

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

• " a point "(x,y)to(y,-x)

$\Rightarrow A \left(7 , 9\right) \to A ' \left(9 , - 7\right) \text{ where A' is the image of A}$

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

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

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

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

$\textcolor{w h i t e}{\Rightarrow \underline{c}} = 2 \left(\begin{matrix}9 \\ - 7\end{matrix}\right) - \left(\begin{matrix}6 \\ 2\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{c}} = \left(\begin{matrix}18 \\ - 14\end{matrix}\right) - \left(\begin{matrix}6 \\ 2\end{matrix}\right) = \left(\begin{matrix}12 \\ - 16\end{matrix}\right)$

$\Rightarrow C = \left(12 , - 16\right)$