# Point A is at (4 ,-2 ) and point B is at (1 ,-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?

The new point $A \left(- 2 , - 4\right)$
and No change in distance. Same distance$= \sqrt{10}$

#### Explanation:

the equation passing thru the origin (0, 0) and point A(4, -2) is

$y = - \frac{1}{2} x$

and distance from (0, 0) to A(4, -2) is $= 2 \sqrt{5}$

the equation passing thru the origin (0, 0) and perpendicular to line $y = - \frac{1}{2} x$ is

$y = 2 x \text{ " }$first equation

We are looking for a new point $A \left({x}_{1} , {y}_{1}\right)$ on the line $y = 2 x$ at the 3rd quadrant and $2 \sqrt{5}$ away from the origin (0, 0).

Use distance formula for "distance from line to a point not on the line"

d=(ax_1+by_1+c)/(+-sqrt(a^2+b^2)

Use equation, $y = - \frac{1}{2} x$ which is also $x + 2 y = 0$

$a = 1$ and $b = 2$ and $c = 0$

$d = 2 \sqrt{5} = \frac{{x}_{1} + 2 {y}_{1}}{\pm \sqrt{{1}^{2} + {2}^{2}}}$

$2 \sqrt{5} = \frac{{x}_{1} + 2 {y}_{1}}{- \sqrt{5}} \text{ " }$choose negative because,
$\left({x}_{1} , {y}_{1}\right)$ is below the line.

$- 2 \sqrt{5} \cdot \sqrt{5} = {x}_{1} + 2 {y}_{1}$

$- 10 = {x}_{1} + 2 {y}_{1} \text{ " }$second equation

${y}_{1} = 2 {x}_{1} \text{ " }$first equation

Using first and second equation. Solve for $\left({x}_{1} , {y}_{1}\right)$

and ${x}_{1} = - 2$ and ${y}_{1} = - 4$