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2013 AMC 12B Problem 8

Problem 8 of 25EasierGeometry

Line ℓ1\ell_1 has equation 3x−2y=13x - 2y = 1 and goes through A=(−1,−2).A = (-1, -2). Line ℓ2\ell_2 has equation y=1y = 1 and meets line ℓ1\ell_1 at point B.B. Line ℓ3\ell_3 has positive slope, goes through point A,A, and meets ℓ2\ell_2 at point C.C. The area of △ABC\triangle ABC is 3.3. What is the slope of ℓ3?\ell_3?

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Solution

Solving 3x−2y=13x - 2y = 1 with y=1y = 1 gives B=(1,1).B = (1, 1). The distance from A=(−1,−2)A = (-1, -2) to the line y=1y = 1 is 3,3, so 12⋅BC⋅3=3\tfrac12\cdot BC\cdot 3 = 3 gives BC=2.BC = 2. Then C=(3,1)C = (3, 1) or C=(−1,1);C = (-1, 1); the latter makes ℓ3\ell_3 vertical, so C=(3,1)C = (3, 1) and the slope is 1−(−2)3−(−1)=34.\dfrac{1 - (-2)}{3 - (-1)} = \dfrac34. Thus, the correct answer is B.

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Concepts: coordinate geometry · triangle area · slope

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