Difference between revisions of "2007 AMC 12A Problems/Problem 13"
(→See also) |
m (wik) |
||
Line 1: | Line 1: | ||
==Problem== | ==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 piece of cheese is located at <math>(12,10)</math> in a [[coordinate plane]]. A mouse is at <math>(4,-2)</math> and is running up the [[line]] <math>y=-5x+18</math>. At the point <math>(a,b)</math> the mouse starts getting farther from the cheese rather than closer to it. What is <math>a+b</math>? |
− | |||
− | |||
− | |||
− | |||
− | |||
− | |||
+ | <math>\mathrm{(A)}\ 6\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 14\qquad \mathrm{(D)}\ 18\qquad \mathrm{(E)}\ 22</math> | ||
==Solution== | ==Solution== | ||
− | We are trying to find the | + | 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 recipricol 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>. |
− | |||
==See also== | ==See also== | ||
− | + | {{AMC12 box|year=2007|ab=A|num-b=12|num-a=14}} | |
− | + | ||
− | + | [[Category:Introductory Geometry Problems]] |
Revision as of 15:28, 10 September 2007
Problem
A piece of cheese is located at in a coordinate plane. A mouse is at and is running up the line . At the point the mouse starts getting farther from the cheese rather than closer to it. What is ?
Solution
We are trying to find the foot of a perpendicular from to . Then the slope of the line that passes through the cheese and is the negative recipricol of the slope of the line, or . Therefore, the line is . The point where and intersect is , and .
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
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 |