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

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<math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math>
 
<math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math>
 
==Solution==
 
 
We are trying to find the
 
 
)<math> to </math>y=-5x+18<math>. Then the [[slope]] of the line that passes through the cheese and </math>(a,b)<math> is the negative reciprocal of the slope of the line, or </math>\frac 15<math>. Therefore, the line is </math>y=\frac{1}{5}x+\frac{38}{5}<math>. The point where </math>y=-5x+18<math> and </math>y=\frac 15x+\frac{38}5<math> intersect is </math>(2,8)<math>, and </math>2+8=10\ (B)$.
 
  
 
==See also==
 
==See also==

Revision as of 07:06, 18 August 2020

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 $a+b$?

$\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions

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