# Difference between revisions of "2007 AMC 12A Problems/Problem 13"

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We are trying to find the point where (12,10) is perpendicular to y=-5x+18. Then the slope of the line that passes through the cheese and (a,b) is 1/5. Therefore, the line is | We are trying to find the point where (12,10) is perpendicular to y=-5x+18. Then the slope of the line that passes through the cheese and (a,b) is 1/5. Therefore, the line is | ||

* <math>y=\frac{1}{5}x+\frac{38}{5}</math> The point where y=-5x+18 and y=.2x+7.6 intersect is (2,8), and 2+8=10(B). | * <math>y=\frac{1}{5}x+\frac{38}{5}</math> The point where y=-5x+18 and y=.2x+7.6 intersect is (2,8), and 2+8=10(B). | ||

+ | |||

+ | ==See also== | ||

+ | * [[2007 AMC 12A Problems/Problem 11 | Previous problem]] | ||

+ | * [[2007 AMC 12A Problems/Problem 13 | Next problem]] | ||

+ | * [[2007 AMC 12A Problems]] |

## Revision as of 08:44, 10 September 2007

## 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?

A(6) B(10) C(14) D(18) E(22)

## Solution

We are trying to find the point where (12,10) is perpendicular to y=-5x+18. Then the slope of the line that passes through the cheese and (a,b) is 1/5. Therefore, the line is

- The point where y=-5x+18 and y=.2x+7.6 intersect is (2,8), and 2+8=10(B).