Point A is at #(1 ,-1 )# and point B is at #(-2 ,9 )#. 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
Yonas Yohannes
Mar 15, 2016

#bar(AB) - bar(A'B) = sqrt(109) -sqrt(73) = 1.8963#

Explanation:

Given point A(1, -1) and point B(-2, 9). Rotate A by #(3pi)/2# and find the difference between #bar(AB)# and #bar(A'B)#
First find #bar(AB)# using the distance formula:
#AB = sqrt((1-(-2))^2 + ((-1)-9)^2) = sqrt(109)#
Now rotate #R_((3pi)/2)(A(1, -1))# this is a rotation by #270^o# clockwise. #A'=R_((3pi)/2)(A(1, -1)) = A'(1,1)#
If you want to show this use:
#A' = ((costheta, sintheta), (-sintheta, costheta)) ((1), (-1))#
#A' = ((0, -1), (1, 0)) ((1), (-1)) = ((1), (1))#
Now compute #bar(A'B) = sqrt((1-(-2))^2 + (1-9)^2) = sqrt(73)#
Now the difference is:
#bar(AB) - bar(A'B) = sqrt(109) -sqrt(73) = 1.8963#

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