# Point A is at (-1 ,-8 ) and point B is at (-5 ,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?

May 21, 2016

(-8 ,1) and ≈ 8.104

#### Explanation:

Under a rotation of $\frac{\pi}{2}$ clockwise about O.

a point (x ,y) → (y ,-x)

hence point A(-1 ,-8) → A' (-8 ,1)

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

length of AB$\left({x}_{1} , {y}_{1}\right) = \left(- 1 , - 8\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 3\right)$

d_(AB)=sqrt((-5+1)^2+(3+8)^2)=sqrt(16+121)≈11.705

length of A'B
$\left({x}_{1} , {y}_{1}\right) = \left(- 8 , 1\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 3\right)$

d_(A'B)=sqrt((-5+8)^2+(3-1)^2)=sqrt(9+4)≈3.601

hence the change in length = 11.705 - 3.601 = 8.104