2017 AMC 10B Problems/Problem 8: Difference between revisions

Latest revision as of 20:00, 21 October 2025

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 left $3$ 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

@@ Line 4: / Line 4: @@
 <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>
-==Solutions==
+==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>.
+<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 <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>.
+==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 <math>B</math> to <math>D</math>, we go to the left <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>
+==Solution 4==
+As stated in solution 1, the triangle is isosceles.
 <asy>
 pair A,B,C,D;
@@ Line 20: / Line 47: @@
 </asy>
-===Solution 1===
+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>.
-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===
-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>.
-===Solution 3===
+- youtube.com/indianmathguy
-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 the <math>D</math>r, 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>
 ==Video Solution==
@@ Line 34: / Line 56: @@
 ~savannahsolver
-==Video Solution 2==
+==Video Solution by TheBeautyofMath==
 https://youtu.be/XRfOULUmWbY?t=367

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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