Question

Which shows the pre-image of triangle X'Y'Z' before the figure was rotated 90° about the origin?

On a coordinate plane, a triangle has points X prime (negative 2, 2), Y prime (0, 6), Z prime (1, 1).

On a coordinate plane, a triangle has points X (2, negative 2), Y (6, 0), Z (1, 1).
On a coordinate plane, a triangle has points X (2, 2), Y (6, 0), and Z (1, negative 1).
On a coordinate plane, a triangle has points X (negative 2, negative 2), Y (6, 0), Z (negative 1, 1).
On a coordinate plane, a triangle has points X (negative 2, 2), Y (negative 6, 0), Z (negative 1, negative 1).

Answer 1

Rotate the image back 90° clockwise (inverse of a 90° counterclockwise rotation). The inverse map is (x',y') -> (y', -x').

Apply to each point:
- X'(-2,2) -> X(2,2)
- Y'(0,6) -> Y(6,0)
- Z'(1,1) -> Z(1,-1)

So the correct pre-image is: X(2,2), Y(6,0), Z(1,-1).