2007 AMC 12A Problems/Problem 13

Revision as of 06:06, 18 August 2020 by Tanish kulkarni (talk | contribs) (Solution)

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

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

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AMC 12 Problems and Solutions

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