# Point A is at (-5 ,1 ) and point B is at (2 ,-3 ). 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
Jun 2, 2017

The distance has changed by $= 4.46$

#### Explanation:

The matrix of a rotation clockwise by $\frac{3}{2} \pi$ about the origin is

$= \left(\begin{matrix}\cos \left(- \frac{3}{2} \pi\right) & - \sin \left(- \frac{3}{2} \pi\right) \\ \sin \left(- \frac{3}{2} \pi\right) & \cos \left(- \frac{3}{2} \pi\right)\end{matrix}\right) = \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right)$

Therefore, the trasformation of point $A$ into $A '$ is

$A ' = \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) \left(\begin{matrix}- 5 \\ 1\end{matrix}\right) = \left(\begin{matrix}- 1 \\ - 5\end{matrix}\right)$

Distance $A B$ is

$= \sqrt{{\left(2 - \left(- 5\right)\right)}^{2} + {\left(- 3 - \left(- 1\right)\right)}^{2}}$

$= \sqrt{49 + 16}$

$= \sqrt{65}$

Distance $A ' B$ is

$= \sqrt{{\left(2 - \left(- 1\right)\right)}^{2} + {\left(- 3 - \left(- 5\right)\right)}^{2}}$

$= \sqrt{9 + 4}$

$= \sqrt{13}$

The distance has changed by

$= \sqrt{65} - \sqrt{13}$

$= 4.46$