# Point A is at (1 ,-1 ) and point B is at (-2 ,9 ). 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?

Mar 15, 2016

$\overline{A B} - \overline{A ' B} = \sqrt{109} - \sqrt{73} = 1.8963$

#### Explanation:

Given point A(1, -1) and point B(-2, 9). Rotate A by $\frac{3 \pi}{2}$ and find the difference between $\overline{A B}$ and $\overline{A ' B}$
First find $\overline{A B}$ using the distance formula:
$A B = \sqrt{{\left(1 - \left(- 2\right)\right)}^{2} + {\left(\left(- 1\right) - 9\right)}^{2}} = \sqrt{109}$
Now rotate ${R}_{\frac{3 \pi}{2}} \left(A \left(1 , - 1\right)\right)$ this is a rotation by ${270}^{o}$ clockwise. $A ' = {R}_{\frac{3 \pi}{2}} \left(A \left(1 , - 1\right)\right) = A ' \left(1 , 1\right)$
If you want to show this use:
$A ' = \left(\begin{matrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{matrix}\right) \left(\begin{matrix}1 \\ - 1\end{matrix}\right)$
$A ' = \left(\begin{matrix}0 & - 1 \\ 1 & 0\end{matrix}\right) \left(\begin{matrix}1 \\ - 1\end{matrix}\right) = \left(\begin{matrix}1 \\ 1\end{matrix}\right)$
Now compute $\overline{A ' B} = \sqrt{{\left(1 - \left(- 2\right)\right)}^{2} + {\left(1 - 9\right)}^{2}} = \sqrt{73}$
Now the difference is:
$\overline{A B} - \overline{A ' B} = \sqrt{109} - \sqrt{73} = 1.8963$