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#