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2007 AMC 12A Problems/Problem 13

Revision as of 14:28, 10 September 2007 by Azjps (talk | contribs) (wik)

Problem

A piece of cheese is located at $(12,10)$ in a coordinate plane. A mouse is at $(4,-2)$ and is running up the line $y=-5x+18$. At the point $(a,b)$ the mouse starts getting farther from the cheese rather than closer to it. What is $a+b$?

$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

Solution

We are trying to find the foot of a perpendicular from $(12,10)$ to $y=-5x+18$. Then the slope of the line that passes through the cheese and $(a,b)$ is the negative recipricol of the slope of the line, or $\frac 15$. Therefore, the line is $y=\frac{1}{5}x+\frac{38}{5}$. The point where $y=-5x+18$ and $y=\frac 15x+\frac{38}5$ intersect is $(2,8)$, and $2+8=10\ (B)$.

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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
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