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?

1 Answer
Dec 21, 2017

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#