# Point A is at (3 ,-2 ) and point B is at (5 ,-4 ). 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?

Apr 16, 2016

(2 , 3) , ≈ 4.79

#### Explanation:

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

a point (x , y) → (-y , x )

hence A (3 , -2) → A' (2 , 3 )

To find the change in distance , requires to calculate the length of AB and A'B , and subtract them to find the change.

We can calculate the lengths 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$

For length AB let $\left({x}_{1} , {y}_{1}\right) = \left(3 , - 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , - 4\right)$

 d_(AB) = sqrt((5-3)^2 + (-4+2)^2)=sqrt(4+4) ≈ 2.83

For A'B let $\left({x}_{1} , {y}_{1}\right) = \left(2 , 3\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(5 , - 4\right)$

 d_(A'B) = sqrt((5-2)^2 + (-4-3)^2)=sqrt(9+49) ≈ 7.62

hence , change in length = 7.62 - 2.83 = 4.79