AMC 12 Mock Exam 3 Problem 8

The midpoint of $\overline{AB}$ is $(4,6)$ and the slope of $\overline{AB}$ is $\frac86=\frac43$, so the bisector is $y-6=-\frac34(x-4)$. Setting $y=0$ gives $x=12$, so $P=(12,0)$. By the shoelace formula, $$[ABP]=\tfrac12\left|1(10-0)+7(0-2)+12(2-10)\right|=\tfrac12\left|10-14-96\right|=\boxed{50}.$$ (Check: $PA=PB=\sqrt{125}$ and $AB=10$, so the height from $P$ is $\sqrt{125-25}=10$ and the area is $\frac12\cdot10\cdot10$.)