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2004 AMC 12A Problem 12

Problem 12 of 25IntermediateGeometry

Let A=(0,9)A = (0, 9) and B=(0,12).B = (0, 12). Points A′A' and B′B' are on the line y=x,y = x, and AA′‾\overline{AA'} and BB′‾\overline{BB'} intersect at C=(2,8).C = (2, 8). What is the length of A′B′‾?\overline{A'B'}?

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Solution

Line ACAC passes through (0,9)(0, 9) with slope 8−92−0=−12,\tfrac{8 - 9}{2 - 0} = -\tfrac12, so its equation is y=−12x+9.y = -\tfrac12 x + 9. Setting y=xy = x gives A′=(6,6).A' = (6, 6). Line BCBC passes through (0,12)(0, 12) with slope −2,-2, so y=−2x+12.y = -2x + 12. Setting y=xy = x gives B′=(4,4).B' = (4, 4). Then A′B′=(6−4)2+(6−4)2=22. \begin{aligned} A'B' &= \sqrt{(6 - 4)^2 + (6 - 4)^2} \\ &= 2\sqrt{2}. \end{aligned} Thus, the correct answer is B.
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