2014 AMC 10A Problem 14

Write $P = (0, p)$ and $Q = (0, -p)$ with $p > 0$; the origin $O$ is the midpoint of $\overline{PQ}$.

Since the lines $AP$ and $AQ$ are perpendicular, $\angle PAQ = 90^\circ$, so $A$ lies on the circle whose diameter is $\overline{PQ}$. That circle is centered at $O$ with radius $p$, hence
$$p = OA = \sqrt{6^2 + 8^2} = 10.$$

Now $\triangle APQ$ has base $PQ = 20$ along the $y$-axis, and its height is the horizontal distance from $A$ to the $y$-axis, which is $6$. Therefore
$$[APQ] = \frac12 \cdot 20 \cdot 6 = 60.$$

The answer is $\boxed{\textbf{(D)}\ 60}$.

"Two perpendicular lines through $A$" and "intercepts symmetric about the origin" together describe a right angle inscribed in a circle with a known center, and Thales' theorem converts that into a single distance computation. Whenever a right angle sits on a segment with a known midpoint, think "circle with that diameter."

Let the slopes be $m$ and $-\frac1m$. The $y$-intercepts are $8 - 6m$ and $8 + \frac6m$; their sum is zero when $16 = 6m - \frac6m$, i.e. $3m^2 - 8m - 3 = 0$, so $m = 3$ or $-\frac13$. Either way the intercepts are $\pm 10$, and the area is $\frac12 \cdot 20 \cdot 6 = 60$.

• Solving for the two slopes with algebra and making a sign error, or using the wrong height (8 instead of the horizontal distance 6) for the triangle.
• Assuming the lines have slopes $\pm 1$ because the intercepts are symmetric, which gives $PQ = 12$ and area $36$.