# Point A is at (6 ,-2 ) and point B is at (-3 ,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?

Sep 28, 2016

Use $d = \sqrt{{\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}}$ to compute the distance, convert A to polar form and rotate $- \frac{\pi}{2}$, convert A back to Cartesian form, compute the new distance, and the difference.

#### Explanation:

Compute the distance from B to A:

$d = \sqrt{{\left(6 - - 3\right)}^{2} + {\left(- 2 - 5\right)}^{2}}$

$d = \sqrt{130}$

Convert A to polar form:

$r = \sqrt{{6}^{2} + {\left(- 2\right)}^{2}}$
$r = \sqrt{40}$
${\theta}_{1} = {\tan}^{-} 1 \left(- \frac{2}{6}\right)$

Rotate:

$r = \sqrt{{6}^{2} + {\left(- 2\right)}^{2}}$
$r = \sqrt{40}$
${\theta}_{2} = {\tan}^{-} 1 \left(- \frac{2}{6}\right) - \frac{\pi}{2}$

Convert back to Cartesian form:

(rcos(theta_2), rsin(theta_2) = (-2, -6)

Compute the distance from B to the new A:

${d}_{2} = \sqrt{{\left(- 2 - - 3\right)}^{2} + {\left(- 6 - 5\right)}^{2}}$

d_2 = sqrt(1 + (-11)^2

$\Delta d = \sqrt{130} - \sqrt{122}$