2013 AMC 12B Problem 8

Attempt Problem 8 of the 2013 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2013 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

8.

Line 1\ell_1 has equation 3x2y=13x - 2y = 1 and goes through A=(1,2).A = (-1, -2). Line 2\ell_2 has equation y=1y = 1 and meets line 1\ell_1 at point B.B. Line 3\ell_3 has positive slope, goes through point A,A, and meets 2\ell_2 at point C.C. The area of ABC\triangle ABC is 3.3. What is the slope of 3?\ell_3?

23\dfrac{2}{3}

34\dfrac{3}{4}

11

43\dfrac{4}{3}

32\dfrac{3}{2}

Answer: B
Concepts:coordinate geometrytriangle areaslope
Difficulty rating: 1460
Solution:

Solving 3x2y=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 12BC3=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 3\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.

← Problem 7#7
Full Exam

Problem 8 in Other Years