Free GMAT Problem Solving Practice Question

This is a reflection problem. The robot must touch the line y = 2 once, and both P and Q lie above it, since their y-coordinates 4 and 9 are both greater than 2. To minimize a path that bounces off a line, reflect one endpoint across the line and draw a straight segment: any path from P to a point R on the line and then to Q has the same length as the path from the reflected point P′ to R to Q, because reflecting does not change the length of the first leg. The shortest such path is the straight line from P′ to Q, which automatically picks the best touch point.

Reflecting P across y = 2 keeps the x-coordinate and sends y to (2 × 2) − y, so P = (1, 4) becomes P′ = (1, 0). The straight-line distance from P′ = (1, 0) to Q = (13, 9) has horizontal change 12 and vertical change 9, so it is √(12² + 9²) = √(144 + 81) = √225 = 15. The segment from P′ to Q crosses y = 2 at an x between 1 and 13, so a valid touch point exists and 15 is actually reached. The answer is C.

Among the wrong choices: choice A (9) sums the two vertical drops to the line, (4 − 2) + (9 − 2) = 9, and forgets that the points are also far apart horizontally. Choice B (13) is the direct P-to-Q distance from someone who never reflects and ignores that the path must touch the line. Choice D (17) adds the horizontal gap to the direct vertical gap, a right-angle reading of the direct route instead of a straight-line one. Choice E (21) is the literal down-across-up path, which does touch the line but is longer than the straight reflected route. The key idea is that the minimum uses the sum of the two heights above the line, 2 + 7 = 9, as the vertical leg of the reflected triangle, not the difference between them.