Question

Point A is at `(-2 ,8 )` and point B is at `(-1 ,3 )`. Point A is rotated `(3pi)/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

The change in distance is:

`Deltad = d_2 - d_1`

`Deltad = sqrt82 - sqrt26`

Explanation:

The first distance from B to A is:

`d_1 = sqrt((-2 - -1)^2 + (8 - 3)^2)`

`d_1 = sqrt((-1)^2 + (5)^2)`

`d_1 = sqrt(1 + 25)`

`d_1 = sqrt26`

Compute the polar coordinates of point A:

`r = sqrt((-2)^2 + (8)^2)`

`r = sqrt(4 + 64)`

`r = sqrt68`

`theta_1 = tan^-1(8/-2) + pi`

Note: We add `pi`, because the inverse tangent function returns either first quadrant angles or negative rotations. In this case, it will return a negative angle.

To rotate the point we add `(3pi)/2` to the original angle:

`theta_2 = tan^-1(8/-2) + (5pi)/2`

Compute the x and y coordinates for new point A:

`(sqrt68cos(tan^-1(8/-2) + (5pi)/2), sqrt68sin(tan^-1(8/-2) + (5pi)/2))`

`(8, 2)`

Compute the new distance:

`d_2 = sqrt((8 - -1)^2 + (2 - 3)^2)`

`d_2 = sqrt((9)^2 + (-1)^2)`

`d_2 = sqrt82`

The change in distance is:

`Deltad = d_2 - d_1`

`Deltad = sqrt82 - sqrt26`