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 $C$ ?

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

Solutions

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

Solution 1

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

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

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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2017 AMC 10B Problems/Problem 8

Contents

Problem

Solutions

Solution 1

Solution 2

See Also