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== | + | |
| + | ==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 <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== | + | ===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 13:51, 13 August 2017
Problem
Points 
and 
are vertices of 
with 
. The altitude from 
meets the opposite side at 
. What are the coordinates of point 
?
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]](http://latex.artofproblemsolving.com/d/6/c/d6c903419860d4d2d6fb832a27d73e8e7acd40ae.png)
Solution 1
Since 
, then is isosceles, so 
. Therefore, the coordinates of are 
.
Solution 2
Calculating the equation of the line running between points 
and 
, 
. The only coordinate of that is also on this line is 
.
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. 
