Question

Point A is at `(8 ,-4 )` and point B is at `(-9 ,6 )`. 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 `=(4,8)` and the distance has changed by `=6.32`

Explanation:

The point `A=(8,-4)`

The point `B=(-9,6)`

The distance

`AB=sqrt((-9-8)^2+(6-(-4))^2)=sqrt(289+100)=sqrt325`

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),(-4))=((4),(8))`

The distance

`A'B=sqrt((-2-9)^2+(8-4)^2)=sqrt(121+16)=sqrt137`

The change in the distance is

`AB-A'B=sqrt325-sqrt137=6.32u`