Point A is at #(3 ,9 )# and point B is at #(-2 ,3 )#. 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
Walker P
Feb 25, 2017

#A = (9, -3)#
#Delta d = sqrt(157)-sqrt(61)#

Explanation:

First let's consider where A and B are on the Cartesian plane:
enter image source here

At the beginning of the problem, we can determine the initial distance between A and B using the distance formula:

#d = sqrt((x_A-x_B)^2+(y_A-y_B)^2)#

Note: This is just a variation of the Pythagorean theorem.

Plugging in the initial coordinates, the original distance between A and B is:
#d = sqrt((3-(-2))^2+(9-3)^2)#
#d = sqrt((5)^2+(6)^2)#
#d = sqrt(25+36)=sqrt(61)#

The problem states that A is rotated by #pi/2#, or #90 deg#, in the clockwise direction. To figure out the new coordinates of A, you can use what is called a "linear transformation". Essentially, if you can determine the new coordinates of #vec i# and #vec j#, you can easily determine the new coordinates of A.

After a #90 deg# rotation, #vec i# and #vec j# will be:

#vec i = (0, -1)#
#vec j = (1, 0)#

Multiplying the original coordinates of A by the new coordinates of #vec i# and #vec j#, we get:

#A = (9, -3)#

Note: If this doesn't make sense, I highly recommend watching the following starting at 3:30 -

Let's visualize this rotation:
enter image source here

Next, calculate the new distance using the distance formula:
#d = sqrt((x_A-x_B)^2+(y_A-y_B)^2)#
#d = sqrt((9-(-2))^2+(-3-3)^2)#
#d = sqrt((11)^2+(-6)^2)#
#d = sqrt(121+36)#
#d = sqrt(157)#

Overall, the change in distance would be:
#Delta d = sqrt(157) - sqrt(61)#

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