2007 AMC 12A Problems/Problem 13

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 reciprocal 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 $2+8=10\ (B)$ .

2007 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

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2007 AMC 12A Problems/Problem 13

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