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Difference between revisions of "2011 AIME I Problems/Problem 3"

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Converting both equations to the form <math>0=Ax+By+C</math>, we have that <math>L</math> has the equation <math>0=12x+5y-90</math> and that <math>M</math> has the equation <math>0=5x-12y-132</math>.
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Converting both equations to the form <math>0=Ax+By+C</math>, we have that <math>L</math> has the equation <math>0=5x-12y-132</math> and that <math>M</math> has the equation <math>0=12x+5y-90</math>.
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Applying the point-to-line distance formula, <math>\frac{\abs{Ax+By+C}}{\sqrt{A^2+B^2}}</math>, to point <math>P</math> and lines <math>L</math> and <math>M</math>, we find that the distance from <math>P</math> to <math>L</math> and <math>M</math> are <math>\frac{123}{13}</math> and <math>\frac{526}{13}</math>, respectively.  
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Applying the point-to-line distance formula, <math>\frac{\abs{Ax+By+C}}{\sqrt{A^2+B^2}}</math>, to point <math>P</math> and lines <math>L</math> and <math>M</math>, we find that the distance from <math>P</math> to <math>L</math> and <math>M</math> are <math>\frac{526}{13}</math> and <math>\frac{123}{13}</math>, respectively.  
  
  

Revision as of 14:36, 14 May 2011

Problem

Let $L$ be the line with slope $\frac{5}{12}$ that contains the point $A=(24,-1)$, and let $M$ be the line perpendicular to line $L$ that contains the point $B=(5,6)$. The original coordinate axes are erased, and line $L$ is made the $x$-axis and line $M$ the $y$-axis. In the new coordinate system, point $A$ is on the positive $x$-axis, and point $B$ is on the positive $y$-axis. The point $P$ with coordinates $(-14,27)$ in the original system has coordinates $(\alpha,\beta)$ in the new coordinate system. Find $\alpha+\beta$.

Solution

Given that $L$ has slope $\frac{5}{12}$ and contains the point $A=(24,-1)$, we may write the point-slope equation for $L$ as $y+1=\frac{5}{12}(x-24)$. Since $M$ is perpendicular to $L$ and contains the point $B=(5,6)$, we have that the slope of $M$ is $-\frac{12}{5}$, and consequently that the point-slope equation for $M$ is $y-6=-\frac{12}{5}(x-5)$.


Converting both equations to the form $0=Ax+By+C$, we have that $L$ has the equation $0=5x-12y-132$ and that $M$ has the equation $0=12x+5y-90$. Applying the point-to-line distance formula, $\frac{\abs{Ax+By+C}}{\sqrt{A^2+B^2}}$ (Error compiling LaTeX. Unknown error_msg), to point $P$ and lines $L$ and $M$, we find that the distance from $P$ to $L$ and $M$ are $\frac{526}{13}$ and $\frac{123}{13}$, respectively.


Since $A$ and $B$ lie on the positive axes of the shifted coordinate plane, we may show by graphing the given system that point P will lie in the second quadrant in the new coordinate system. Therefore, the abscissa of $P$ is negative, and is therefore $-\frac{123}{13}$; similarly, the ordinate of $P$ is positive and is therefore $\frac{526}{13}$.

Thus, we have that $\alpha=-\frac{123}{13}$ and that $\beta=\frac{526}{13}$. It follows that $\alpha+\beta=-\frac{123}{13}+\frac{526}{13}=\frac{403}{13}=\boxed{031}$.

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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