# Point A is at (3 ,-2 ) and point B is at (2 ,1 ). 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 14, 2016

≈ 1.937

#### Explanation:

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

a point (x , y) → (-x , -y)

hence A(3 , -2) → (-3 , 2)

Now , we have to calculate the difference between 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$\left({x}_{1} , {y}_{1}\right) \text{ and "(x_2,y_2)" are 2 coordinate points }$

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

 d_(AB) = sqrt((2-3)^2 + (1+2)^2) = sqrt(1+9) = sqrt10 ≈ 3.162

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

d_(A'B) = sqrt((2+3)^2 + (1-2)^2) = sqrt(25+1)=sqrt26 ≈ 5.099

difference = A'B - AB = 5.099 - 3.162 = 1.937