# Point A is at (3 ,9 ) and point B is at (-2 ,3 ). 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 25, 2017

$A = \left(9 , - 3\right)$
$\Delta d = \sqrt{157} - \sqrt{61}$

#### Explanation:

First let's consider where A and B are on the Cartesian plane:

At the beginning of the problem, we can determine the initial distance between A and B using the distance formula:

$d = \sqrt{{\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}}$

Note: This is just a variation of the Pythagorean theorem.

Plugging in the initial coordinates, the original distance between A and B is:
$d = \sqrt{{\left(3 - \left(- 2\right)\right)}^{2} + {\left(9 - 3\right)}^{2}}$
$d = \sqrt{{\left(5\right)}^{2} + {\left(6\right)}^{2}}$
$d = \sqrt{25 + 36} = \sqrt{61}$

The problem states that A is rotated by $\frac{\pi}{2}$, or $90 \mathrm{de} g$, in the clockwise direction. To figure out the new coordinates of A, you can use what is called a "linear transformation". Essentially, if you can determine the new coordinates of $\vec{i}$ and $\vec{j}$, you can easily determine the new coordinates of A.

After a $90 \mathrm{de} g$ rotation, $\vec{i}$ and $\vec{j}$ will be:

$\vec{i} = \left(0 , - 1\right)$
$\vec{j} = \left(1 , 0\right)$

Multiplying the original coordinates of A by the new coordinates of $\vec{i}$ and $\vec{j}$, we get:

$A = \left(9 , - 3\right)$

Note: If this doesn't make sense, I highly recommend watching the following starting at 3:30 -

Let's visualize this rotation:

Next, calculate the new distance using the distance formula:
$d = \sqrt{{\left({x}_{A} - {x}_{B}\right)}^{2} + {\left({y}_{A} - {y}_{B}\right)}^{2}}$
$d = \sqrt{{\left(9 - \left(- 2\right)\right)}^{2} + {\left(- 3 - 3\right)}^{2}}$
$d = \sqrt{{\left(11\right)}^{2} + {\left(- 6\right)}^{2}}$
$d = \sqrt{121 + 36}$
$d = \sqrt{157}$

Overall, the change in distance would be:
$\Delta d = \sqrt{157} - \sqrt{61}$