Question

Point A is at `(6 ,-2 )` and point B is at `(-3 ,5 )`. 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

Use `d = sqrt((x_A - x_B)^2 + (y_A - y_B)^2)` to compute the distance, convert A to polar form and rotate `-pi/2`, convert A back to Cartesian form, compute the new distance, and the difference.

Explanation:

Compute the distance from B to A:

`d = sqrt((6 - -3)^2 + (-2 - 5)^2)`

`d = sqrt(130)`

Convert A to polar form:

`r = sqrt(6^2 + (-2)^2)`
`r = sqrt(40)`
`theta_1 = tan^-1(-2/6)`

Rotate:

`r = sqrt(6^2 + (-2)^2)`
`r = sqrt(40)`
`theta_2 = tan^-1(-2/6) - pi/2`

Convert back to Cartesian form:

`(rcos(theta_2), rsin(theta_2) = (-2, -6)`

Compute the distance from B to the new A:

`d_2 = sqrt((-2 - -3)^2 + (-6 - 5)^2)`

`d_2 = sqrt(1 + (-11)^2`

`Deltad = sqrt(130) - sqrt(122)`