Sample Question 5:

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 \(b-a?\)

\(\textbf{(A) }1 \qquad \textbf{(B) }3 \qquad \textbf{(C) }5 \qquad \textbf{(D) }7 \qquad \textbf{(E) }9\)

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Answer Keys

Question 5: D

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Solutions

Question 5
Step 1: 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 \((-3,6)\).

Step 2: 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 \(4\).

Step 3: 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 \(P\).

Step 4: Therefore point \(P\) is located at \((1+1, 5+4) = (2,9)\). The answer is \(9-2 = 7\), so the correct answer choice is \(\textbf{(D)} ~7\).