2009
Problem - 727
Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$?
The length of $AB$ is $3\sqrt{2}$. Meanwhile, $AB$ is parallel to $x+y=7$ with a distance of $2\sqrt{2}$. Hence the area of $\triangle{ABC}$ will be $\boxed{6}$.