Difference between revisions of "2011 AIME I Problems/Problem 3"
Flamefoxx99 (talk | contribs) m (→Solution) |
|||
Line 10: | Line 10: | ||
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>. | 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>. | ||
− | Applying the point-to-line distance formula, <math>\frac{ | + | Applying the point-to-line distance formula, <math>\frac{|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 22:01, 1 March 2015
Problem
Let be the line with slope
that contains the point
, and let
be the line perpendicular to line
that contains the point
. The original coordinate axes are erased, and line
is made the
-axis and line
the
-axis. In the new coordinate system, point
is on the positive
-axis, and point
is on the positive
-axis. The point
with coordinates
in the original system has coordinates
in the new coordinate system. Find
.
Solution
Given that has slope
and contains the point
, we may write the point-slope equation for
as
.
Since
is perpendicular to
and contains the point
, we have that the slope of
is
, and consequently that the point-slope equation for
is
.
Converting both equations to the form , we have that
has the equation
and that
has the equation
.
Applying the point-to-line distance formula,
, to point
and lines
and
, we find that the distance from
to
and
are
and
, respectively.
Since and
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
is negative, and is therefore
; similarly, the ordinate of
is positive and is therefore
.
Thus, we have that and that
. It follows that
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.