Point A is at (9 ,-2 ) and point B is at (1 ,-3 ). Point A is rotated pi/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 25, 2016

(-2 ,-9) , ≈ 1.13

Explanation:

Under a rotation of $\frac{\pi}{2} \text{ clockwise about 0 }$

a point (x ,y) → (y ,-x)

hence A (9 ,-2) → A' (-2 , -9)

To calculate the change in distance we need to find the length of AB and A'B 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 $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 points }$

calculate AB

let $\left({x}_{1} , {y}_{1}\right) = \left(9 , - 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , - 3\right)$

d=sqrt((1-9)^2+(-3+2)^2)=sqrt(64+1)=sqrt65 ≈ 8.06

calculate A'B

let $\left({x}_{1} , {y}_{1}\right) = \left(- 2 , - 9\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , - 3\right)$

d=sqrt((1+2)^2+(-3+9)^2)=sqrt(9+36)=sqrt45 ≈ 6.93

change in length = 8.06 - 6.93 = 1.13