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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