# Point A is at (-1 ,-3 ) and point B is at (-5 ,4 ). 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?

Feb 7, 2018

$\left(- 3 , 1\right) , \approx 4.457$

#### Explanation:

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

• " a point "(x,y)to(y,-x)

$\Rightarrow A \left(- 1 , - 3\right) \to A ' \left(- 3 , 1\right) , \text{A' is the image of A}$

$\text{to calculate the difference in distances use the "color(blue)"distance formula}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{let "(x_1,y_1)=(-1,-3)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 4\right)$

$A B = \sqrt{{\left(- 5 + 1\right)}^{2} + {\left(4 + 3\right)}^{2}} = \sqrt{16 + 49} = \sqrt{65}$

$\text{let "(x_1,y_1)=(-3,1)" and } \left({x}_{2} , {y}_{2}\right) = \left(- 5 , 4\right)$

$A ' B = \sqrt{{\left(4 - 1\right)}^{2} + {\left(- 5 + 3\right)}^{2}} = \sqrt{9 + 4} = \sqrt{13}$

$\text{change in distance "=sqrt65-sqrt13~~4.457" 3 dec. places}$