Point A is at (5 ,-2 ) and point B is at (-2 ,5 ). 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?

Nov 24, 2017

See below.

Explanation:

It can be seen from the diagram, that a rotation about the origin through an angle $\theta$ can be represented as:

$\left(\begin{matrix}1 \\ 0\end{matrix}\right) \to \left(\begin{matrix}\cos \theta \\ \sin \theta\end{matrix}\right)$ and $\left(\begin{matrix}0 \\ 1\end{matrix}\right) \to \left(\begin{matrix}- \sin \theta \\ \textcolor{w h i t e}{8} \cos \theta\end{matrix}\right)$

So the transformation matrix will be:

$\left(\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \textcolor{w h i t e}{8} \cos \theta\end{matrix}\right)$

Matrix for point A:

$A = \left(\begin{matrix}5 \\ - 2\end{matrix}\right)$

Transformation:- Rotation through $\frac{\pi}{2}$ clockwise. This is equivalent to $2 \pi - \frac{\pi}{2} = \frac{3 \pi}{2}$ anticlocwise.

$\left(\begin{matrix}\cos \left(\frac{3 \pi}{2}\right) & - \sin \left(\frac{3 \pi}{2}\right) \\ \sin \left(\frac{3 \pi}{2}\right) & \textcolor{w h i t e}{8} \cos \left(\frac{3 \pi}{2}\right)\end{matrix}\right) \left(\begin{matrix}\textcolor{w h i t e}{88} 5 \\ - 2\end{matrix}\right) = \left(\begin{matrix}- 2 \\ \textcolor{w h i t e}{} - 5\end{matrix}\right)$

$A ' = \left(\begin{matrix}- 2 \\ \textcolor{w h i t e}{} - 5\end{matrix}\right)$

Distance between A and B:

$d = \sqrt{{\left(5 - \left(- 2\right)\right)}^{2} + {\left(- 2 - \left(- 5\right)\right)}^{2}} = 7 \sqrt{2}$

Distance between A' and B:

$d = \sqrt{{\left(- 2 - \left(- 2\right)\right)}^{2} + {\left(- 5 - 5\right)}^{2}} = 10$

The distance between the points has increased by a factor of $\frac{5 \sqrt{2}}{7}$