# Point A is at (-2 ,8 ) and point B is at (-1 ,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?

Sep 30, 2016

The change in distance is:

$\Delta d = {d}_{2} - {d}_{1}$

$\Delta d = \sqrt{82} - \sqrt{26}$

#### Explanation:

The first distance from B to A is:

${d}_{1} = \sqrt{{\left(- 2 - - 1\right)}^{2} + {\left(8 - 3\right)}^{2}}$

${d}_{1} = \sqrt{{\left(- 1\right)}^{2} + {\left(5\right)}^{2}}$

${d}_{1} = \sqrt{1 + 25}$

${d}_{1} = \sqrt{26}$

Compute the polar coordinates of point A:

$r = \sqrt{{\left(- 2\right)}^{2} + {\left(8\right)}^{2}}$

$r = \sqrt{4 + 64}$

$r = \sqrt{68}$

${\theta}_{1} = {\tan}^{-} 1 \left(\frac{8}{-} 2\right) + \pi$

Note: We add $\pi$, because the inverse tangent function returns either first quadrant angles or negative rotations. In this case, it will return a negative angle.

To rotate the point we add $\frac{3 \pi}{2}$ to the original angle:

${\theta}_{2} = {\tan}^{-} 1 \left(\frac{8}{-} 2\right) + \frac{5 \pi}{2}$

Compute the x and y coordinates for new point A:

$\left(\sqrt{68} \cos \left({\tan}^{-} 1 \left(\frac{8}{-} 2\right) + \frac{5 \pi}{2}\right) , \sqrt{68} \sin \left({\tan}^{-} 1 \left(\frac{8}{-} 2\right) + \frac{5 \pi}{2}\right)\right)$

$\left(8 , 2\right)$

Compute the new distance:

${d}_{2} = \sqrt{{\left(8 - - 1\right)}^{2} + {\left(2 - 3\right)}^{2}}$

${d}_{2} = \sqrt{{\left(9\right)}^{2} + {\left(- 1\right)}^{2}}$

${d}_{2} = \sqrt{82}$

The change in distance is:

$\Delta d = {d}_{2} - {d}_{1}$

$\Delta d = \sqrt{82} - \sqrt{26}$