2009 AMC 12B Problems/Problem 9

Problem

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 $\triangle ABC$ ?

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 $\frac 12 \cdot 4 \cdot 3 = \boxed {6}$ .

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 $\mathrm{(A)}$ .