# Point A is at (1 ,3 ) and point B is at (-7 ,-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 16, 2018

color(blue)((3,-1)

$\textcolor{b l u e}{0.54337889 \setminus \setminus u n i t s}$

#### Explanation:

We can produce a rotation about the origin by using the transformation matrix:

$\left(\begin{matrix}\cos \left(\theta\right) & - \sin \left(\theta\right) \\ \sin \left(\theta\right) & \cos \left(\theta\right)\end{matrix}\right)$

This is for anticlockwise rotation, so for clockwise rotation we use the angle:

$2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$

So we have:

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

$A = \left(1 , 3\right)$

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

Distance between A and B:

$d = \sqrt{{\left(1 - \left(- 7\right)\right)}^{2} + {\left(3 - \left(- 5\right)\right)}^{2}} = \sqrt{128} = 8 \sqrt{2}$

Distance between A' and B:

$d = \sqrt{{\left(3 - \left(- 7\right)\right)}^{2} + {\left(- 1 - \left(- 5\right)\right)}^{2}} = \sqrt{116} = 2 \sqrt{29}$

Change in distance:

$8 \sqrt{2} - 2 \sqrt{29} = 0.54337889$units