Point A is at (2 ,-1 ) and point B is at (3 ,-4 ). Point A is rotated pi 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 31, 2016

|bar(AB) - bar(A'B)| = sqrt(26) - 5sqrt(2)|

Explanation:

Given : Two Points A(2, -1) and B(3, -4), Rotate A by pi
Required: New coordinates of A(2, -1) =>_(R(pi)) A'(x, y) and distance |bar(AB) - bar(A'B)|
Solution strategy :
a) Rotate A
b) Use distance formula to find AB and A'B
c) Calculate |bar(AB) - bar(A'B)|

color(red)a) A' = R(pi)A where R(pi) is the 2xx2 rotation matrix
R(pi)=[(costheta, -sintheta ), (sintheta, costheta)] _(theta=pi)

R(pi)=[(1, 0), (0, 1)] so then,

A' = [(-1, 0), (0, -1)][(2), (-1)] = [(-2), (1)]
that is A'(-2,1)
You can also get this geometrically or by inspection...

color(red)b) bar(AB)= sqrt((2-3)^2 + (-1-4)^2) = sqrt(26)
bar(A'B) = sqrt((-2-3)^2 + (1-(-4))^2) = 5sqrt(2)

|bar(AB) - bar(A'B)| = sqrt(26) - 5sqrt(2)|

Related questions
See all questions in Rotations
Impact of this question
1600 views around the world
You can reuse this answer
Creative Commons License