2014 AMC 10A Problem 14

We have that the $y$-intercepts are an equal distance from the origin since their values sum to $0.$ Let this distance be $z.$ Because the two given lines are perpendicular, $\triangle APQ$ is right at $A$. The origin is the midpoint of its hypotenuse $PQ$, so it is equidistant from $P$, $Q$, and $A$. Hence the distance from $A$ to the origin is also $z$. We then know that $$z = \sqrt{6^2 + 8^2} = 10$$ by the distance formula. We know the altitude from $A$ to $\overline{PQ}$ is $6$ (it is just the $x$-value of $A$). We also know that $PQ = 2 \cdot 10 = 20,$ which tells us that the area $$[APQ] = \dfrac{1}{2} \cdot 6 \cdot 20 = 60.$$ Thus, D is the correct answer.