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?

1 Answer
Dec 23, 2017

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#