Difference between revisions of "2018 AMC 12B Problems/Problem 3"
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Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10} | Using the slope-intercept form, we get the equations <math>y-30 = 6(x-40)</math> and <math>y-30 = 2(x-40)</math>. Simplifying, we get <math>6x-y=210</math> and <math>2x-y=50</math>. Letting <math>y=0</math> in both equations and solving for <math>x</math> gives the <math>x</math>-intercepts: <math>x=35</math> and <math>x=25</math>, respectively. Thus the distance between them is <math>35-25 = 10 \Rightarrow \boxed{(\text{B}) 10} | ||
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==See Also== | ==See Also== | ||
Revision as of 01:09, 19 February 2018
Problem
A line with slope 2 intersects a line with slope 6 at the point 
. What is the distance between the 
-intercepts of these two lines?
Solution 1
Using the slope-intercept form, we get the equations 
and 
. Simplifying, we get 
and 
. Letting 
in both equations and solving for gives the -intercepts: 
and 
, respectively. Thus the distance between them is 
See Also
| 2018 AMC 12B (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
