# Point A is at (-3 ,4 ) and point B is at (-8 ,1 ). Point A is rotated (3pi)/2  clockwise about the origin. What are the new coordinates of point A and by how much has the distance between points A and B changed?

##### 1 Answer
Nov 24, 2017

It can be seen from the diagram, that a rotation about the origin through an angle $\theta$ can be represented as:

$\left(\begin{matrix}1 \\ 0\end{matrix}\right) \to \left(\begin{matrix}\cos \theta \\ \sin \theta\end{matrix}\right)$ and $\left(\begin{matrix}0 \\ 1\end{matrix}\right) \to \left(\begin{matrix}- \sin \theta \\ \textcolor{w h i t e}{8} \cos \theta\end{matrix}\right)$

So the transformation matrix will be:

$\left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \textcolor{w h i t e}{8} \cos \theta\end{matrix}\right)$

Matrix A:

$A = \left(\begin{matrix}- 3 \\ \textcolor{w h i t e}{8} 4\end{matrix}\right)$

Transformation matrix will be:

$\left(\begin{matrix}\cos \left(\frac{3 \pi}{2}\right) & - \sin \left(\frac{3 \pi}{2}\right) \\ \sin \left(\frac{3 \pi}{2}\right) & \textcolor{w h i t e}{8} \cos \left(\frac{3 \pi}{2}\right)\end{matrix}\right)$
$\therefore$

$A ' = \left(\begin{matrix}\cos \left(\frac{3 \pi}{2}\right) & - \sin \left(\frac{3 \pi}{2}\right) \\ \sin \left(\frac{3 \pi}{2}\right) & \textcolor{w h i t e}{8} \cos \left(\frac{3 \pi}{2}\right)\end{matrix}\right) \left(\begin{matrix}- 3 \\ \textcolor{w h i t e}{8} 4\end{matrix}\right) = \left(\begin{matrix}4 \\ 3\end{matrix}\right)$

Coordinates:

$\left(4 , 3\right)$

Distance between A and B:

$d = \sqrt{{\left(- 3 - \left(- 8\right)\right)}^{2} + {\left(4 - 1\right)}^{2}} = \sqrt{34}$

Distance between $A ' \mathmr{and} B$

$d = \sqrt{{\left(4 - \left(- 8\right)\right)}^{2} + {\left(3 - 1\right)}^{2}} = 4$

The distance has been reduced by a factor of $\frac{2 \sqrt{34}}{17}$

$\therefore$

$\frac{2 \sqrt{34}}{17} \cdot \sqrt{34} = 4$