If $a,b>0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by=6$ has area 6, then $ab=$
$\mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }$
Setting $y=0$ we have that the $x-$intercept of the line is $x= \frac{6}{a}$. Similarly setting $x=0$ we find the $y-$intercept to be $y= \frac{6}{b}$. Then $\frac{18}{ab}=\frac{1}{2}\cdot\frac{6}{a}\cdot\frac{6}{b}$ so that $\frac{18}{ab} = 6$, simplifying we would get $ab=3$. Hence the answer is $\fbox{A}$.
-$\LaTeX$ by Kevinliu08