Difference between revisions of "2003 AMC 10B Problems/Problem 11"
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If an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope <math>5</math>, we find | If an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope <math>5</math>, we find | ||
<cmath>\frac{\text{change in }y{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_2}=5\implies x_2=-3</cmath>. | <cmath>\frac{\text{change in }y{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_2}=5\implies x_2=-3</cmath>. | ||
| − | Therefore, the <math>x</math>-intercept of the line with slope <math>5</math> is <math>(10-3,0)=(7,0)</math>. The distance between | + | Therefore, the <math>x</math>-intercept of the line with slope <math>5</math> is <math>(10-3,0)=(7,0)</math>. The distance between <math>\boxed{(5,0)--(7,0)=2\implies A}</math>. |
==See Also== | ==See Also== | ||
Revision as of 08:42, 6 July 2019
Problem
A line with slope 
intersects a line with slope 
at point 
. What is the distance between the 
-intercepts of these two lines?
Solution 1
Using the point-slope form, the equation of each line is
![\[y-15=3(x-10) \longrightarrow y=3x-15\]](http://latex.artofproblemsolving.com/4/b/0/4b037a952efd3a1d90f21cfae3201b8b27e6e249.png)
![\[y-15=5(x-10) \longrightarrow y=5x-35\]](http://latex.artofproblemsolving.com/9/3/6/9363e785bc07ef6d3345ffff2b3cd345be5fdb2e.png)
Substitute in 
to find the -intercepts.
![\[0=3x-15\longrightarrow x=5\]](http://latex.artofproblemsolving.com/c/8/5/c85ab0e6b9308cb0549a47dac89e43636d46944a.png)
![\[0=5x-35\longrightarrow x=7\]](http://latex.artofproblemsolving.com/7/d/9/7d97578c90c79935d44e9e03ca1693dce68b80d0.png)
The difference between them is 
.
Solution 2(using slopes)
The -intercepts of a line is where the line's 
-coordinate is 
. The slope is defined as 
. Therefore, the line with slope can be expressed as
![\[\frac{\text{change in }y{-coordinate}}{\text{change in }x\text{-coordinate}}=3\]](http://latex.artofproblemsolving.com/0/d/a/0da746244106d1f11149908ef3e197c4f8d0cfa4.png)
From to the -intercept, the line must move 
units to the -axis. Therefore, ![\[\frac{\text{change in }y{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_1}=3\implies x_1=-5\]](http://latex.artofproblemsolving.com/d/0/0/d005d4b0b7cbb10344b5ec5b531e0f93e7044760.png)
.
Therefore, the -intercept of the line with slope is 
.Note that some sources may state "the -intercept is " instead of "the -intercept is 
.
If an idea functions great for one part of the problem but does not find the solution to the problem, we want to use the idea again. Using the idea for the line with slope , we find
![\[\frac{\text{change in }y{-coordinate}}{\text{change in }x\text{-coordinate}}=\frac{-15}{x_2}=5\implies x_2=-3\]](http://latex.artofproblemsolving.com/b/2/a/b2ae5603f3d4d05eb4f618a7f7c57e6bccf57f15.png)
.
Therefore, the -intercept of the line with slope is 
. The distance between 
.
See Also
| 2003 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
