Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 3.5 AMC 10/12 Find the answer

The point $P(a,b)$ in the $xy$ -plane is first rotated counterclockwise by $90^\circ$ around the point $(1,5)$ and then reflected about the line $y = -x$ . The image of $P$ after these two transformations is at $(-6,3)$ . What is $b - a ?$
$\textbf{(A)} ~1$ $\textbf{(B)} ~3$ $\textbf{(C)} ~5$ $\textbf{(D)} ~7$ $\textbf{(E)} ~9$

Your answer

Multiple choice: answer with the letter of the option you want.

Solution

The final image of $P$ is $(-6,3)$ . We know the reflection rule for reflecting over $y=-x$ is $(x,y) \rightarrow (-y, -x)$ . So before the reflection and after rotation the point is $(-3,6)$ .
By definition of rotation, the slope between $(-3,6)$ and $(1,5)$ must be perpendicular to the slope between $(a,b)$ and $(1,5)$ . The first slope is $\frac{5-6}{1-(-3)} = \frac{-1}{4}$ . This means the slope of $P$ and $(1,5)$ is $4$ .
Rotations also preserve distance to the center of rotation, and since we only "travelled" up and down by the slope once to get from $(3,-6)$ to $(1,5)$ it follows we shall only use the slope once to travel from $(1,5)$ to $P$ .
Therefore point $P$ is located at $(1+1, 5+4) = (2,9)$ . The answer is $9-2 = 7 = \boxed{\textbf{(D)} ~7}$ .
-abhinavg0627

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.