# Point A is at (-7 ,3 ) and point B is at (5 ,4 ). 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?

Jun 2, 2017

$A ' = \left(7 , - 3\right)$
The distance has decreased by $\sqrt{145} - \sqrt{53} \approx 4.76$

#### Explanation:

Given: $A \left(- 7 , 3\right) , B \left(5 , 4\right)$ Point $A$ is rotated $\pi$ clockwise about the origin.

A $\pi$ rotation clockwise is a ${180}^{\circ}$ rotation CW.

The coordinate rule for a ${180}^{\circ}$ rotation CW is: $\left(x , y\right) \to \left(- x , - y\right)$

Using the coordinate rule: $A ' = \left(7 , - 3\right)$

The distance formula is $d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$

${d}_{A B} = \sqrt{{\left(5 - - 7\right)}^{2} + {\left(4 - 3\right)}^{2}} = \sqrt{{12}^{2} + {1}^{2}} = \sqrt{145}$

${d}_{A ' B} = \sqrt{{\left(5 - 7\right)}^{2} + {\left(4 - - 3\right)}^{2}} = \sqrt{{2}^{2} + {7}^{2}} = \sqrt{53}$

The distance between point $A$ and $B$ has not changed.

The distance between the rotated point $A$ = $A '$ and $B$ is $\sqrt{145} - \sqrt{53} \approx 4.76$