2017 AMC 10B Problem 8

The triangle has two equal sides, $AB = AC$, so it is isosceles with top vertex $A$. That tells us where the altitude from $A$ lands.

1. In an isosceles triangle, the altitude from the top vertex hits the base exactly at its middle. (The two halves of the triangle are mirror images.) So $D$ is the midpoint of $BC$.
2. Find the step from $B$ to $D$: the $x$-coordinate goes from $2$ to $-1$ (change $-3$), and the $y$-coordinate goes from $-3$ to $3$ (change $+6$).
3. Since $D$ is halfway, $C$ is the same step again from $D$:
$$C = (-1 - 3,\ 3 + 6) = (-4, 9).$$
4. Check that $AB = AC$. From $A(11, 9)$ to $B(2, -3)$ the changes are $9$ and $12$, so $AB = \sqrt{81 + 144} = 15$. From $A(11, 9)$ to $C(-4, 9)$ the change is $15$ across and $0$ up, so $AC = 15$.

The answer is $\boxed{\textbf{(C)}\ (-4, 9)}$.

In an isosceles triangle, the altitude from the top vertex is also the line to the middle of the base. Turning that fact into coordinates makes the problem a midpoint question: going from $B$ to $D$ and then the same amount again reaches $C$. The distance check at the end is quick and catches sign mistakes.

Use the choices. We need $AC = AB = 15$. Choice (C), $(-4, 9)$, has the same $y$-coordinate as $A(11, 9)$ and is $15$ units to its left, so it is exactly $15$ away. You can confirm it also fits $D$ being the midpoint of $B$ and $C$: $\frac{2 + (-4)}{2} = -1$ and $\frac{-3 + 9}{2} = 3$.

Because $AB = AC$, the altitude from the apex $A$ to base $BC$ is also the perpendicular bisector of $BC$. Therefore $D$ is the midpoint of $\overline{BC}$.

Going from $B$ to $D$ moves $(-1 - 2,\ 3 - (-3)) = (-3, 6)$. Moving the same amount again from $D$ reaches $C$:
$$C = (-1 - 3,\ 3 + 6) = (-4, 9).$$

Quick check: $AB^2 = 9^2 + 12^2 = 225$ and $AC^2 = 15^2 + 0^2 = 225$, so $AB = AC$ indeed.

The answer is $\boxed{\textbf{(C)}\ (-4, 9)}$.