2021 AMC 10B Problem 9

Work backwards from the final image, undoing the last transformation first.

Undo the reflection. Reflection over $y = -x$ sends $(x, y) \mapsto (-y, -x)$ and is its own inverse. The point that reflects to $(-6, 3)$ is $(-3, 6)$. This is the rotated image of $P$.

Undo the rotation. To undo a $90^\circ$ counterclockwise rotation about $(1,5)$, rotate $90^\circ$ clockwise about $(1,5)$. The offset of $(-3,6)$ from the center is $(-4, 1)$. A clockwise quarter turn sends an offset $(u, v)$ to $(v, -u)$, so $(-4, 1) \mapsto (1, 4)$. Adding the center back: $P = (1 + 1, 5 + 4) = (2, 9)$.

So $a = 2$, $b = 9$, and $b - a = 7$.

Check forward: offset of $(2,9)$ is $(1,4)$; counterclockwise gives $(-4, 1)$, i.e. the point $(-3, 6)$; reflecting over $y=-x$ gives $(-6, 3)$. Correct.

The answer is $\boxed{\textbf{(D)}\ 7}$.

A composition of transformations is inverted by inverting each step in the opposite order. Rotations about a point other than the origin are handled by subtracting the center, rotating the offset with the standard rule ($(u,v)\mapsto(-v,u)$ counterclockwise), and adding the center back. Always finish with a forward check; it costs seconds and catches sign slips.

• Reflecting over y = x (swapping coordinates) instead of y = −x, or undoing the rotation counterclockwise instead of clockwise.
• Rotating about the origin instead of about $(1,5)$, or forgetting to add the center back after rotating the offset.