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

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==Solution==
 
==Solution==
  
We are trying to find the foot of a [[perpendicular]] from <math>(12,10)</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)</math>.
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We are trying to find the  
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)<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 06:06, 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 $a+b$?

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

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
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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