# 2013 AMC 12B Problems/Problem 8

## Problem

Line $l_1$ has equation $3x - 2y = 1$ and goes through $A = (-1, -2)$. Line $l_2$ has equation $y = 1$ and meets line $l_1$ at point $B$. Line $l_3$ has positive slope, goes through point $A$, and meets $l_2$ at point $C$. The area of $\triangle ABC$ is $3$. What is the slope of $l_3$? $\textbf{(A)}\ \frac{2}{3} \qquad \textbf{(B)}\ \frac{3}{4} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ \frac{4}{3} \qquad \textbf{(E)}\ \frac{3}{2}$

## Solution 1

Line $l_1$ has the equation $y=3x/2-1/2$ when rearranged. Substituting $1$ for $y$, we find that line $l_2$ will meet this line at point $(1,1)$, which is point $B$. We call $\overline{BC}$ the base and the altitude from $A$ to the line connecting $B$ and $C$, $y=-1$, the height. The altitude has length $|-2-1|=3$, and the area of $\triangle{ABC}=3$. Since $A={bh}/2$, $b=2$. Because $l_3$ has positive slope, it will meet $l_2$ to the right of $B$, and the point $2$ to the right of $B$ is $(3,1)$. $l_3$ passes through $(-1,-2)$ and $(3,1)$, and thus has slope $\frac{|1-(-2)|}{|3-(-1)|}=$ $\boxed{\textbf{(B) }\frac{3}{4}}$.

## Solution 2 - Shoelace Theorem

We know lines $l_1$ and $l_2$ intersect at $B$, so we can solve for that point: $$3x-2y=1$$ Because $y = 1$ we have: $$3x-2(1) = 1$$ $$3x-2=1$$ $$3x=3$$ $$x = 1$$

Thus we have $B = (1,1)$.

We know that the area of the triangle is $3$, so by Shoelace Theorem we have: $$A = \dfrac{1}{2} |(-2x+y-1) - (-2+x-y)|$$ $$A = \dfrac{1}{2} |-2x+y-1+2-x+y|$$ $$3 = \dfrac{1}{2} |-2x+y-1+2-x+y|$$ $$6 = |-3x+2y+1|.$$

Thus we have two options: $$6 = -3x+2y+1$$ $$5 = -3x+2y$$

or $$6 = 3x-2y-1$$ $$7 = 3x-2y.$$

Now we must just find a point that satisfies $m_{l_3}$ is positive.

Doing some guess-and-check yields, from the second equation: $$7 = 3x-2y$$ $$7 = 3(3)-2(1)$$ $$7 = 7$$

so a valid point here is $(3,1)$. When calculated, the slope of $l_3$ in this situation yields $\boxed{\textbf{(B) }\frac{3}{4}}$.

## Video Solution

~Punxsutawney Phil or sugar_rush

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