Difference between revisions of "2007 AMC 12A Problems/Problem 13"
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==Solution== | ==Solution== | ||
| − | We are trying to find the foot of a [[perpendicular]] from <math>(12,10)</math> to <math>y=-5x+18</math>. Then the [[slope]] of the line that passes through the cheese and <math>(a,b)</math> is the negative | + | We are trying to find the foot of a [[perpendicular]] from <math>(12,10)</math> to <math>y=-5x+18</math>. Then the [[slope]] of the line that passes through the cheese and <math>(a,b)</math> is the negative reciprocal of the slope of the line, or <math>\frac 15</math>. Therefore, the line is <math>y=\frac{1}{5}x+\frac{38}{5}</math>. The point where <math>y=-5x+18</math> and <math>y=\frac 15x+\frac{38}5</math> intersect is <math>(2,8)</math>, and <math>2+8=10\ (B)</math>. |
==See also== | ==See also== | ||
Revision as of 11:18, 8 December 2011
Problem
A piece of cheese is located at 
in a coordinate plane. A mouse is at 
and is running up the line 
. At the point 
the mouse starts getting farther from the cheese rather than closer to it. What is 
?
Solution
We are trying to find the foot of a perpendicular from to . Then the slope of the line that passes through the cheese and is the negative reciprocal of the slope of the line, or 
. Therefore, the line is 
. The point where and 
intersect is 
, and 
.
See also
| 2007 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 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 12 Problems and Solutions | |
