# Point A is at (2 ,-1 ) and point B is at (3 ,-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?

Mar 31, 2016

$| \overline{A B} - \overline{A ' B} | = \sqrt{26} - 5 \sqrt{2} |$

#### Explanation:

Given : Two Points $A \left(2 , - 1\right)$ and $B \left(3 , - 4\right)$, Rotate $A$ by $\pi$
Required: New coordinates of $A \left(2 , - 1\right) {\implies}_{R \left(\pi\right)} A ' \left(x , y\right)$ and distance $| \overline{A B} - \overline{A ' B} |$
Solution strategy :
a) Rotate A
b) Use distance formula to find $A B \mathmr{and} A ' B$
c) Calculate $| \overline{A B} - \overline{A ' B} |$

color(red)a) $A ' = R \left(\pi\right) A$ where $R \left(\pi\right)$ is the $2 \times 2$ rotation matrix
$R \left(\pi\right) = {\left[\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right]}_{\theta = \pi}$

$R \left(\pi\right) = \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]$ so then,

$A ' = \left[\begin{matrix}- 1 & 0 \\ 0 & - 1\end{matrix}\right] \left[\begin{matrix}2 \\ - 1\end{matrix}\right] = \left[\begin{matrix}- 2 \\ 1\end{matrix}\right]$
that is $A ' \left(- 2 , 1\right)$
You can also get this geometrically or by inspection...

color(red)b) $\overline{A B} = \sqrt{{\left(2 - 3\right)}^{2} + {\left(- 1 - 4\right)}^{2}} = \sqrt{26}$
$\overline{A ' B} = \sqrt{{\left(- 2 - 3\right)}^{2} + {\left(1 - \left(- 4\right)\right)}^{2}} = 5 \sqrt{2}$

$| \overline{A B} - \overline{A ' B} | = \sqrt{26} - 5 \sqrt{2} |$