# 2017 AMC 10B Problems/Problem 8

## Problem

Points $A(11, 9)$ and $B(2, -3)$ are vertices of $\triangle ABC$ with $AB=AC$. The altitude from $A$ meets the opposite side at $D(-1, 3)$. What are the coordinates of point $C$?

$\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)$

## Solution 1

Since $AB = AC$, then $\triangle ABC$ is isosceles, so $BD = CD$. Therefore, the coordinates of $C$ are $(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}$.

$[asy] pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("A",A,E); label("B",B,S); label("D",D,W); label("C",C,N); [/asy]$

## Solution 2

Calculating the equation of the line running between points $B$ and $D$, $y = -2x + 1$. The only coordinate of $C$ that is also on this line is $\boxed{\textbf{(C) } (-4,9)}$.

## Solution 3

Similar to the first solution, because the triangle is isosceles, then the line drawn in the middle separates the triangle into two smaller congruent triangles. To get from $B$ to $D$, we go to the right $3$ and up $6$. Then to get to point $C$ from point $D$, we go to the right $3$ and up $6$, getting us the coordinates $\boxed{\textbf{(C) } (-4,9)}$. ~$\text{KLBBC}$

## Solution 4

As stated in solution 1, the triangle is isosceles.

$[asy] pair A,B,C,D; A=(11,9); B=(2,-3); C=(-4,9); D=(-1,3); draw(A--B--C--cycle); draw(A--D); draw(rightanglemark(A,D,B)); label("A",A,E); label("B",B,S); label("D",D,W); label("C",C,N); [/asy]$

This means that $D(-1, 3)$ is the midpoint of $B(2, -3)$ and $C(x,y)$. So $\frac{x+2}{2}$ $= -1$ and so $x = -4$. Similarly for $y$, we have $\frac{y-3}{2}$ $=3$ and so $y=9$. So our final answer is $\boxed{\textbf{(C) } (-4,9)}$.

~savannahsolver

~IceMatrix