# Points A and B are at (1 ,1 ) and (4 ,6 ), 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?

Jan 26, 2018

The point $C$ is at $\left(- 2 , - 8\right)$.

#### Explanation:

First, rotate $A$ around the origin like the problem describes. The rotation angle in degrees is:

(3cancel(pi))/2*(180º)/cancel(pi) =

3/2*180º = 270º

A 270º rotation is three-fourths of a circle, so the point $A \left(1 , 1\right)$ becomes $A ' \left(1 , - 1\right)$.

Now, we can figure out the equation on the line between $A '$ and $B$ using the slope and point-slope formulae:

$m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \implies$

$m = \frac{- 1 - 6}{1 - 4} = \frac{- 7}{- 3} = \frac{7}{3}$

$y - {y}_{1} = m \left(x - {x}_{1}\right) \implies$

$y - 6 = \frac{7}{3} \left(x - 4\right)$

$y - 6 = \frac{7}{3} x - \frac{28}{3}$

$y = \frac{7}{3} x - \frac{10}{3}$

Since the dilation from point $C$ was a factor of two, we know that the change in $x$ from $A '$ to $B$ is equal to the change in $x$ from $C$ to $A '$.

The change in $x$ from $A '$ to $B$ is $| 1 - 4 | = 3$
We now know that the $x$-coordinate of point $C$ is $x$-coordinate of point $A '$ - 3, or $1 - 3 = - 2$ .

We can plug in -2 to our line that we found to get the $y$-coordinate:

$y = \frac{7}{3} x - \frac{10}{3} \implies$

$y = \frac{7}{3} \left(- 2\right) - \frac{10}{3}$

$y = \frac{- 14}{3} - \frac{10}{3}$

$y = - \frac{24}{3}$

$y = - 8$

Therefore, the point $C$ is $\left(- 2 , - 8\right)$.

Jan 26, 2018

$C = \left(- 2 , - 8\right)$

#### Explanation:

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

• " a point "(x,y)to(y,-x)

$\Rightarrow a \left(1 , 1\right) \to A ' \left(1 , - 1\right) \text{ where A' is the image of A}$

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

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

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

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

$\textcolor{w h i t e}{\Rightarrow \underline{c}} = 2 \left(\begin{matrix}1 \\ - 1\end{matrix}\right) - \left(\begin{matrix}4 \\ 6\end{matrix}\right)$

$\textcolor{w h i t e}{\Rightarrow \underline{c}} = \left(\begin{matrix}2 \\ - 2\end{matrix}\right) - \left(\begin{matrix}4 \\ 6\end{matrix}\right) = \left(\begin{matrix}- 2 \\ - 8\end{matrix}\right)$

$\Rightarrow C = \left(- 2 , - 8\right)$