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== | ==Solution== | ||
| − | + | 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>. | |
| + | |||
==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}} | ||
Revision as of 12:50, 16 February 2017
Problem
Points 
and 
are vertices of 
with 
. The altitude from 
meets the opposite side at 
. What are the coordinates of point 
?
Solution
Since 
, then is isosceles, so 
. Therefore, the coordinates of are 
.
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. 
