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 $\triangle ABC$?

$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 8\qquad \mathrm{(C)}\ 10\qquad \mathrm{(D)}\ 12\qquad \mathrm{(E)}\ 14$
Solution 1
Because the line $x + y = 7$ is parallel to $\overline {AB}$, the area of $\triangle ABC$ is independent of the location of $C$ on that line.  Therefore it may be assumed that $C$ is $(7,0)$.  In that case the triangle has base $AC = 4$ and altitude $3$, so its area is $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$.

Solution 2
The base of the triangle is $AB = \sqrt{3^2 + 3^2} = 3\sqrt 2$.  Its altitude is the distance between the point $A$ and the parallel line $x + y = 7$, which is $\frac 4{\sqrt 2} = 2\sqrt 2$. Therefore its area is $\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}$.  The answer is $\mathrm{(A)}$.


Solution 3
By Shoelace, our area is:
\[\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.\]
We know $x+y=7$ so we get:
\[\frac {1}{2} \cdot |9-21|=\boxed 6\]

Solution 4
WLOG, let the coordinates of $C$ be $(3,4)$ , or any coordinate, for that matter. Applying the shoelace formula, we get the area as $\boxed 6$.
~coolmath2017