Question

Point A is at `(2 ,-1 )` and point B is at `(3 ,-4 )`. Point A is rotated `pi` 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(26) - 5sqrt(2)|`

Explanation:

Given : Two Points `A(2, -1)` and `B(3, -4)`, Rotate `A` by `pi`
Required: New coordinates of `A(2, -1) =>_(R(pi)) A'(x, y)` and distance `|bar(AB) - bar(A'B)|`
Solution strategy :
a) Rotate A
b) Use distance formula to find `AB and A'B`
c) Calculate `|bar(AB) - bar(A'B)|`

`color(red)a)` `A' = R(pi)A` where `R(pi)` is the `2xx2` rotation matrix
`R(pi)=[(costheta, -sintheta ), (sintheta, costheta)] _(theta=pi)`

`R(pi)=[(1, 0), (0, 1)]` so then,

`A' = [(-1, 0), (0, -1)][(2), (-1)] = [(-2), (1)]`
that is `A'(-2,1)`
You can also get this geometrically or by inspection...

`color(red)b)` `bar(AB)= sqrt((2-3)^2 + (-1-4)^2) = sqrt(26)`
`bar(A'B) = sqrt((-2-3)^2 + (1-(-4))^2) = 5sqrt(2)`

`|bar(AB) - bar(A'B)| = sqrt(26) - 5sqrt(2)|`