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Line $l\_1$ has equation $2x-y=3$ and goes through $A=(1,-1)$. Line $l\_2$ has equation $y=1$ and meets line $l\_1$ at point $B$. Line $l\_3$ has negative slope, goes through point $A$, and meets $l\_2$ at point $C$. The area of $\triangle A B C$ is $3$. What is the slope of $l\_3$? (Adapted From 2013 AMC 12B Problems, Question \#8) 
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Line $l\_1$ has the equation $y=2x-3$ when rearranged. Substituting $1$ for $y$, we find that line $l\_2$ will meet this line at point $(2,1)$, which is point $B$. We call $\overline{B C}$ the base and the altitude from $A$ to the line connecting $B$ and $C, y=1$, the height. The altitude has length $\textbar-1-1\textbar=2$. The area of $\triangle A B C=3$. Since $A=\frac{bh}{2}, b=3$. Points that are on the line $y= 1$ and has a distance of $3$ from $B$ are $(5,1)$ and $(-1,1)$. Since $l\_3$ has negative slope, point $C$ is $(-1,1)$. $l\_3$ passes through $(-1,1)$ and $(1,-1)$, and thus has slope $\frac{1-(-1)}{-1-1}=(\mathbf{C}) -1$. 