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?

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Answer 1 · 2017-12-21T14:16:57

The new coordinates are #=(4,8)# and the distance has changed by #=6.32#

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#