Difference between revisions of "2017 AMC 10B Problems/Problem 8"

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<math>\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)</math>
 
<math>\textbf{(A)}\ (-8, 9)\qquad\textbf{(B)}\ (-4, 8)\qquad\textbf{(C)}\ (-4, 9)\qquad\textbf{(D)}\ (-2, 3)\qquad\textbf{(E)}\ (-1, 0)</math>
==Solution 1==
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==Solutions==
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<asy>
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pair A,B,C,D;
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A=(11,9);
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B=(2,-3);
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C=(-4,9);
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D=(-1,3);
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draw(A--B--C--cycle);
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draw(A--D);
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draw(rightanglemark(A,D,B));
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label("$A$",A,E);
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label("$B$",B,S);
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label("$D$",D,W);
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label("$C$",C,N);
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</asy>
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===Solution 1===
 
Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>.
 
Since <math>AB = AC</math>, then <math>\triangle ABC</math> is isosceles, so <math>BD = CD</math>. Therefore, the coordinates of <math>C</math> are <math>(-1 - 3, 3 + 6) = \boxed{\textbf{(C) } (-4,9)}</math>.
  
==Solution 2==
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===Solution 2===
 
Calculating the equation of the line running between points <math>B</math> and <math>D</math>, <math>y = -2x + 1</math>. The only coordinate of <math>C</math> that is also on this line is <math>\boxed{\textbf{(C) } (-4,9)}</math>.
 
Calculating the equation of the line running between points <math>B</math> and <math>D</math>, <math>y = -2x + 1</math>. The only coordinate of <math>C</math> that is also on this line is <math>\boxed{\textbf{(C) } (-4,9)}</math>.
  

Revision as of 14:51, 13 August 2017

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

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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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