# 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?

Dec 21, 2017

The new coordinates are $= \left(1 , - 5\right)$ and the distance has changed by $= 5.04 u$

#### Explanation:

The point $A = \left(- 5 , - 1\right)$

The point $B = \left(2 , - 3\right)$

The distance

$A B = \sqrt{{\left(2 - \left(- 5\right)\right)}^{2} + {\left(- 3 - \left(- 1\right)\right)}^{2}} = \sqrt{49 + 4} = \sqrt{53}$

The matrix for a rotation of angle $\theta$ is

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

And when $\theta = - \frac{3}{2} \pi$

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

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

Therefore,

The coordinates of the point $A '$ after the rotation of the point $A$ clockwise by $\frac{3}{2} \pi$ is

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) \cdot \left(\begin{matrix}- 5 \\ - 1\end{matrix}\right) = \left(\begin{matrix}1 \\ - 5\end{matrix}\right)$

The distance

$A ' B = \sqrt{{\left(2 - 1\right)}^{2} + {\left(- 3 - \left(- 5\right)\right)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$

The change in the distance is

$A B - A ' B = \sqrt{53} - \sqrt{5} = 5.04 u$