Point A is at #(6 ,2 )# and point B is at #(3 ,-8 )#. Point A is rotated #pi/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
Feb 21, 2018

The new point #A'# is #(2,-6)#, and the distance changed by 8.204 units.

Explanation:

There is a formal method of doing this, and there is an easier way for simpler problems. I present the formal method first.

Given a point #(x,y)#, rotating it around the origin by angle #theta# results in the coordinate #(x',y')#:
#x' = x cos theta - y sin theta#
#y' = y cos theta + x sin theta#

Now, imagining you haven't taken trig yet (this is overkill for a 90 degree turn anyways), here's a (perhaps) more intuitive method.

Imagine taking the entire coordinate axis and rotating it 90 degrees clockwise about the origin in your head. The positive x-axis is now where the negative y-axis used to be. The positive y-axis is now where the positive x-axis used to be, and so on.

The point used to be at #(6,2)#. It used to be 6 units to the right and 2 units up. However, since you turned it, it's now 6 units down and 2 units right. Make sure you can visualize this in your head, it's a useful skill. It might help to physically draw the coordinate axis, draw the point, and rotate the paper.

Down means negative y-axis, right means positive x-axis. The new coordinates for point #A'# is #(2, -6)#.

To determine how much the distance changed, we simply take the distances between #A# and #B#, and #A'# and #B#.

#d(A,B) = sqrt((6-3)^2+(2-(-8))^2) = sqrt(109) approx 10.440#
#d(A',B) = sqrt((2-3)^2+(-6-(-8))^2) = sqrt(5) approx 2.236#
#|d(A,B) - d(A',B)| = 8.204#

#therefore# the new coordinates for point #A'# is #(2, -6)# and the distance between points #A# and #B# changed by #8.204# units.

#square#