# Point A is at (1 ,-8 ) and point B is at (-3 ,-2 ). 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?

May 7, 2017

The new coordinates are $= \left(8 , 1\right)$ and the change in the distance is $= 4.2$

#### 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 $A$ is

$A ' = \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) \left(\begin{matrix}1 \\ - 8\end{matrix}\right) = \left(\begin{matrix}8 \\ 1\end{matrix}\right)$

Distance $A B$ is

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

$= \sqrt{16 + 36}$

$= \sqrt{52}$

Distance $A ' B$ is

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

$= \sqrt{121 + 9}$

$= \sqrt{130}$

The distance has changed by

$= \sqrt{130} - \sqrt{52}$

$= 4.2$