# Point A is at (3 ,9 ) and point B is at (-2 ,5 ). Point A is rotated pi/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 29, 2017

The new coordinates of $A = \left(9 , - 3\right)$ and the change in the distance is $= 7.2$

#### Explanation:

The matrix of a rotation clockwise by $\frac{1}{2} \pi$ about the origin is

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

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

$A ' = \left(\begin{matrix}0 & 1 \\ - 1 & 0\end{matrix}\right) \left(\begin{matrix}3 \\ 9\end{matrix}\right) = \left(\begin{matrix}9 \\ - 3\end{matrix}\right)$

Distance $A B$ is

$= \sqrt{{\left(- 2 - 3\right)}^{2} + {\left(5 - \left(9\right)\right)}^{2}}$

$= \sqrt{25 + 16}$

$= \sqrt{41}$

Distance $A ' B$ is

$= \sqrt{{\left(- 2 - 9\right)}^{2} + {\left(5 - \left(- 3\right)\right)}^{2}}$

$= \sqrt{121 + 64}$

$= \sqrt{185}$

The distance has changed by

$= \sqrt{185} - \sqrt{41}$

$= 7.2$