# Point A is at (-7 ,7 ) and point B is at (5 ,-3 ). Point A is rotated pi  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?

Apr 5, 2016

A' = (7,-7) , d ≈ 10.85

#### Explanation:

Under a rotation of $\pi \text{ about the origin }$

a point A (x,y) → A' (-x,-y)

hence A (-7,7) → A' (7,-7)

To calculate the change in distance we will need to calculate the distance from A to B and also the distance from A' to B.

Using the $\textcolor{b l u e}{\text{ distance formula }}$

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

where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$

distance between A and B

here $\left({x}_{1} , {y}_{1}\right) = \left(- 7 , 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , - 3\right)$

d =sqrt((5-(-7))^2+(-3-7)^2) = sqrt(144+100) ≈ 15.62

distance between A' and B

here $\left({x}_{1} , {y}_{1}\right) = \left(7 , - 7\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , - 3\right)$

d = sqrt((5-7)^2+(-3+7)^2) = sqrt(4+16) ≈ 4.77

change in distance = 15.62 - 4.77 = 10.85