Question

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

$\dfrac{2}{3}$

$\dfrac{3}{4}$

$1$

$\dfrac{4}{3}$

$\dfrac{3}{2}$

Solution:

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

2013 AMC 12B Problem 8

Solution:

Problem 8 in Other Years