Question

Point A is at `(1 ,-1 )` and point B is at `(-2 ,9 )`. 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

`bar(AB) - bar(A'B) = sqrt(109) -sqrt(73) = 1.8963`

Explanation:

Given point A(1, -1) and point B(-2, 9). Rotate A by `(3pi)/2` and find the difference between `bar(AB)` and `bar(A'B)`
First find `bar(AB)` using the distance formula:
`AB = sqrt((1-(-2))^2 + ((-1)-9)^2) = sqrt(109)`
Now rotate `R_((3pi)/2)(A(1, -1))` this is a rotation by `270^o` clockwise. `A'=R_((3pi)/2)(A(1, -1)) = A'(1,1)`
If you want to show this use:
`A' = ((costheta, sintheta), (-sintheta, costheta)) ((1), (-1))`
`A' = ((0, -1), (1, 0)) ((1), (-1)) = ((1), (1))`
Now compute `bar(A'B) = sqrt((1-(-2))^2 + (1-9)^2) = sqrt(73)`
Now the difference is:
`bar(AB) - bar(A'B) = sqrt(109) -sqrt(73) = 1.8963`