Point A is at (4 ,-9 ) and point B is at (-2 ,8 ). 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?

1 Answer
Dec 20, 2017

The new coordinates are =(9,4). The distance has changed by =6.32u

Explanation:

The point A=(4,-9)

The point B=(-2,8)

The distance

AB=sqrt((-2-4)^2+(8-(-9))^2)=sqrt(36+289)=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))*((4),(-9))=((9),(4))

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