Question

Point A is at `(8 ,-1 )` and point B is at `(9 ,-7 )`. 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 `=(1,8)` and the distace has changed by `=11.9u`

Explanation:

The point `A=(8,-1)`

The point `B=(9,-7)`

The distance

`AB=sqrt((9-8)^2+(-7-(-1))^2)=sqrt(1+25)=sqrt26`

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))*((8),(-1))=((1),(8))`

The distance

`A'B=sqrt((9-1)^2+(-7-8)^2)=sqrt(64+225)=sqrt289=17`

The change in the distance is

`AB-A'B=17-sqrt26=11.9u`