Question

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?

Answer 1

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

Explanation:

First let's consider where A and B are on the Cartesian plane:

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:

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)`