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).
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==See also==
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* [[2007 AMC 12A Problems/Problem 11 | Previous problem]]
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* [[2007 AMC 12A Problems/Problem 13 | Next problem]]
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* [[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

  • $y=\frac{1}{5}x+\frac{38}{5}$ The point where y=-5x+18 and y=.2x+7.6 intersect is (2,8), and 2+8=10(B).

See also

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