# Point A is at (5 ,-8 ) and point B is at (-3 ,3 ). 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 27, 2018

See below.

#### Explanation:

A rotation of $\frac{\pi}{2}$ clockwise maps:

$\left(x , y\right) \to \left(- y , x\right)$

Points: $A = \left(5 , - 8\right)$, $\left(- 3 , 3\right)$

$\therefore$

$A ' = \left(5 , - 8\right) \to \left(- y , x\right) = \left(8 , 5\right)$

$B ' = \left(- 3 , 3\right) \to \left(- y , x\right) = \left(- 3 , - 3\right)$

Where $A ' \mathmr{and} B '$ are the images of A and B respectively.

Using the distance formula to find the distance between A and B and A' and B'

$A \mathmr{and} B$

$d = \sqrt{{\left(- 3 - 5\right)}^{2} + \left(3 - \left(- 8\right)\right)} = \sqrt{185}$

$A ' \mathmr{and} B '$

$d = \sqrt{{\left(- 3 - 8\right)}^{2} + {\left(- 3 , - 5\right)}^{2}} = \sqrt{185}$

We didn't need to even calculate this. A rotation doesn't change the relative distance between points. The question is very ambiguous when it asks for distance between A and B. I have assumed it meant between $A ' \mathmr{and} B '$.