Line $l\_1$ has the equation $y=\frac{3x}{2}-\frac12$ 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 $\overline{B C}$ the base and the altitude from $A$ to the line connecting $B$ and $C, y=1$, the height. The altitude has length $\textbar-2-1\textbar=3$, and the area of $\triangle A B C=3$. Since $A=\frac{bh}{2}, b=2$. Because $l\_3$ has positive slope, it will meet $l\_2$ to the right of $B$, and the point that is $2$ to the right of $B$ is $(3,1)$. $l\_3$ passes through $(-1,-2)$ and $(3,1)$, and thus has slope $\frac{\textbar1-(-2)\textbar}{\textbar3-(-1)\textbar}=(\mathbf{B}) \frac{3}{4}$. 
Line $l\_1$ has equation $3 x-2 y=1$ 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 $\triangle A B C$ is $3$ . What is the slope of $l\_3$? (2013 AMC 12B Problems, Question \#8) 

* $$\frac23$$ 
* $$\frac34$$ 
* $$1$$ 
* $$\frac43$$ 
* $$\frac32$$ 