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AMC12 2021b Test Paper
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Sample Question 5:

The point P(a,b)P(a,b) in the xyxy-plane is first rotated counterclockwise by 9090^\circ around the point (1,5)(1,5) and then reflected about the line y=xy=-x. The image of PP after these two transformations is at (6,3)(-6,3). What is ba?b-a?

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




Answer Keys

Question 5: D


Solutions

Question 5
Step 1: The final image of PP is (6,3)(-6,3). We know the reflection rule for reflecting over y=xy=-x is (x,y)(y,x)(x,y) \rightarrow (-y, -x). So before the reflection and after rotation, the point is (3,6)(-3,6).

Step 2: By definition of rotation, the slope between (3,6)(-3,6) and (1,5)(1,5) must be perpendicular to the slope between (a,b)(a,b) and (1,5)(1,5). The first slope is 561(3)=14\frac{5-6}{1-(-3)} = \frac{-1}{4}. This means the slope of PP and (1,5)(1,5) is 44.

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)(3,-6) to (1,5)(1,5) it follows we shall only use the slope once to travel from (1,5)(1,5) to PP.

Step 4: Therefore point PP is located at (1+1,5+4)=(2,9)(1+1, 5+4) = (2,9). The answer is 92=79-2 = 7, so the correct answer choice is (D) 7\textbf{(D)} ~7.