Question

Point A is at `(-5 ,-1 )` and point B is at `(2 ,-3 )`. Point A is rotated `(3pi)/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?

Answer 1

The new coordinates are `=(1,-5)` and the distance has changed by `=5.04u`

Explanation:

The point `A=(-5,-1)`

The point `B=(2,-3)`

The distance

`AB=sqrt((2-(-5))^2+(-3-(-1))^2)=sqrt(49+4)=sqrt53`

The matrix for a rotation of angle `theta` is

`r(theta)=((costheta,-sintheta),(sintheta,costheta))`

And when `theta=-3/2pi`

`r(-3/2pi)=((cos(-3/2pi),-sin(-3/2pi)),(sin(-3/2pi),cos(-3/2pi)))`

`=((0,-1),(1,0))`

Therefore,

The coordinates of the point `A'` after the rotation of the point `A` clockwise by `3/2pi` is

`((x),(y))=((0,-1),(1,0))*((-5),(-1))=((1),(-5))`

The distance

`A'B=sqrt((2-1)^2+(-3-(-5))^2)=sqrt(1+4)=sqrt5`

The change in the distance is

`AB-A'B=sqrt53-sqrt5=5.04u`