# Point A is at (4 ,1 ) and point B is at (-9 ,-7 ). 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?

May 19, 2016

(-1 ,4) and ≈ 1.664

#### Explanation:

Under a rotation of $\frac{3 \pi}{2} \text{ clockwise about the origin}$

A point (x ,y) → (-y ,x)

Hence point A(4 ,1) → A'(-1 ,4)

We now require to calculate the distance between A and B and between A' and B.
We can do this using the $\textcolor{b l u e}{\text{ distance formula}}$

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points}$

For A to B let$\left({x}_{1} , {y}_{1}\right) = \left(4 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 9 , - 7\right)$

d=sqrt((-9-4)^2+(-7-1)^2)=sqrt(169+64)≈15.264

For A' to B$\left({x}_{1} , {y}_{1}\right) = \left(- 1 , 4\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 9 , - 7\right)$

d=sqrt((-9+1)^2+(-7-4)^2)=sqrt(64+121)≈13.60

The difference in distance = 15.264 - 13.60 = 1.664