# Difference between revisions of "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 $2+8=10\ (B)$.