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== |
| + | 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> | <asy> | ||
pair A,B,C,D; | pair A,B,C,D; | ||
| Line 20: | Line 22: | ||
</asy> | </asy> | ||
| − | + | ==Solution 2== | |
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| − | |||
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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>. | 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 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> | 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> | ||
Revision as of 17:37, 17 January 2021
Contents
Problem
Points 
and 
are vertices of 
with 
. The altitude from 
meets the opposite side at 
. What are the coordinates of point 
?
Solution 1
Since 
, then is isosceles, so 
. Therefore, the coordinates of are 
.
![[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]](http://latex.artofproblemsolving.com/d/6/c/d6c903419860d4d2d6fb832a27d73e8e7acd40ae.png)
Solution 2
Calculating the equation of the line running between points 
and 
, 
. The only coordinate of that is also on this line is 
.
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 to the r, we go to the right 
and up 
. Then to get to point from point , we go to the right and up , getting us the coordinates . ~
Video Solution
~savannahsolver
Video Solution 2
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
