Difference between revisions of "2017 AMC 10B Problems/Problem 8"
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draw(A--B--C--cycle); | draw(A--B--C--cycle); | ||
draw(A--D); | draw(A--D); | ||
| − | draw(rightanglemark(A,D,B)); | + | draw(rightanglemark(A,D,B,30)); |
label("$A$",A,E); | label("$A$",A,E); | ||
label("$B$",B,S); | label("$B$",B,S); | ||
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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>. | 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>. | ||
| − | - | + | - youtube.com/indianmathguy |
==Video Solution== | ==Video Solution== | ||
Latest revision as of 16:14, 13 September 2024
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,30)); label("$A$",A,E); label("$B$",B,S); label("$D$",D,W); label("$C$",C,N); [/asy]](http://latex.artofproblemsolving.com/4/e/6/4e64b7d6c733817509b22e4cdd26b7aa91ed42a2.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 , 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 . ~
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]](http://latex.artofproblemsolving.com/d/6/c/d6c903419860d4d2d6fb832a27d73e8e7acd40ae.png)
This means that is the midpoint of and 
. So 
and so 
. Similarly for 
, we have 
and so 
. So our final answer is .
- youtube.com/indianmathguy
Video Solution
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
