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

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==Problem==
 
==Problem==
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Placeholder
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Points <math>A(11, 9)</math> and <math>B(2, -3)</math> are vertices of <math>\triangle ABC</math> with <math>AB=AC</math>. The altitude from <math>A</math> meets the opposite side at <math>D(-1, 3)</math>. What are the coordinates of point <math>C</math>?
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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>
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==Solution 1==
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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>.
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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,30));
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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 2==
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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>.
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==Solution 3==
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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 <math>B</math> to <math>D</math>, we go to the right <math>3</math> and up <math>6</math>. Then to get to point <math>C</math> from point <math>D</math>, we go to the right <math>3</math> and up <math>6</math>, getting us the coordinates <math>\boxed{\textbf{(C) } (-4,9)}</math>. ~<math>\text{KLBBC}</math>
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==Solution 4==
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As stated in solution 1, the triangle is isosceles.
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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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This means that <math>D(-1, 3)</math> is the midpoint of <math>B(2, -3)</math> and <math>C(x,y)</math>. So <math>\frac{x+2}{2}</math> <math>= -1</math> and so <math>x = -4</math>. Similarly for <math>y</math>, we have <math>\frac{y-3}{2}</math> <math>=3</math> and so <math>y=9</math>. So our final answer is <math>\boxed{\textbf{(C) } (-4,9)}</math>.
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- youtube.com/indianmathguy
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==Video Solution==
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https://youtu.be/4rRckA3gcPU
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~savannahsolver
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==Video Solution by TheBeautyofMath==
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https://youtu.be/XRfOULUmWbY?t=367
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 +
~IceMatrix
  
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==Solution==
 
−
Placeholder
 
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=7|num-a=9}}
 
{{AMC10 box|year=2017|ab=B|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 16:14, 13 September 2024

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

- youtube.com/indianmathguy

Video Solution

https://youtu.be/4rRckA3gcPU

~savannahsolver

Video Solution by TheBeautyofMath

https://youtu.be/XRfOULUmWbY?t=367

~IceMatrix

See Also

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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