# Point A is at (-2 ,-8 ) and point B is at (-5 ,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?

Mar 12, 2016

Let Initial polar coordinate of A ,$\left(r , \theta\right)$
Given Initial Cartesian coordinate of A ,$\left({x}_{1} = - 2 , {y}_{1} = - 8\right)$
So we can write
$\left({x}_{1} = - 2 = r \cos \theta \mathmr{and} {y}_{1} = - 8 = r \sin \theta\right)$
After $3 \frac{\pi}{2}$ clockwise rotation the new coordinate of A becomes
${x}_{2} = r \cos \left(- 3 \frac{\pi}{2} + \theta\right) = r \cos \left(3 \frac{\pi}{2} - \theta\right) = - r \sin \theta = - \left(- 8\right) = 8$

${y}_{2} = r \sin \left(- 3 \frac{\pi}{2} + \theta\right) = - r \sin \left(3 \frac{\pi}{2} - \theta\right) = r \cos \theta = - 2$

Initial distance of A from B(-5,3)
${d}_{1} = \sqrt{{3}^{2} + {11}^{2}} = \sqrt{130}$
final distance between new position of A(8,-2) and B(-5,3)
${d}_{2} = \sqrt{{13}^{2} + {5}^{2}} = \sqrt{194}$
So Difference=$\sqrt{194} - \sqrt{130}$