# Point A is at (1 ,-4 ) and point B is at (-9 ,-2 ). 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 11, 2016

hew coordinate (4,1)
Change in distance $\Delta = {d}_{2} - {d}_{1} = \sqrt{178} - \sqrt{104}$

#### Explanation:

Distance of initial position of the point A i.e. P (1.-4) from origin O is the radius of the circular path $r$ = $\sqrt{{\left(1 - 0\right)}^{2} + {\left(- 4 - 0\right)}^{2}} = \sqrt{17}$ not required

If the initial $\angle X O P = - \theta$
Initial x-coordinate of A at P is ${x}_{1} = 1 = r \cos \left(- \theta\right) \implies r \cos \left(\theta\right) = 1$
Initial y-coordinate of A at P is ${y}_{1} = = - 4 = r \sin \left(- \theta\right) \implies r \sin \left(\theta\right) = 4$

After clockwise rotation at an angle $3 \frac{\pi}{2} = {270}^{o}$ in a circular path of radius  the point takes new position Q having coordinate $\left({x}_{2} , {y}_{2}\right)$
${x}_{2} = r \cos \left(- 270 - \theta\right) = r \cos \left(270 + \theta\right) = r \sin \theta = 4$
${y}_{2} = r \sin \left(- 270 - \theta\right) = - r \sin \left(270 + \theta\right) = r \cos \theta = 1$

The distance between P,Q
$d = \sqrt{{\left(1 - 4\right)}^{2} + {\left(- 4 - 1\right)}^{2}} = \sqrt{34}$ not wanted
Initial distance between A(1,-4)i.e.P andB(-9-2)
${d}_{1} = \sqrt{{\left(1 + 9\right)}^{2} + {\left(- 4 + 2\right)}^{2}} = \sqrt{104}$

Fnitial distance between A(4,1)i.e.Q andB(-9-2)
${d}_{2} = \sqrt{{\left(4 + 9\right)}^{2} + {\left(1 + 2\right)}^{2}} = \sqrt{178}$

Change in distance $\Delta = {d}_{2} - {d}_{1} = \sqrt{178} - \sqrt{104}$