Difference between revisions of "2018 AMC 12B Problems/Problem 3"
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− | ==Problem== | + | == Problem == |
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A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? | A line with slope 2 intersects a line with slope 6 at the point <math>(40,30)</math>. What is the distance between the <math>x</math>-intercepts of these two lines? | ||
<math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math> | <math>(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50</math> | ||
− | + | == Solutions == | |
− | ==Solution 1== | + | === Solution 1 === |
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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} | ||
\indent</math> | \indent</math> | ||
− | ==Solution 2== | + | === Solution 2 === |
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In order for the line with slope <math>2</math> to travel "up" <math>30</math> units (from <math>y=0</math>), it must have traveled <math>30/2=15</math> units to the right. Thus, the <math>x</math>-intercept is at <math>x=40-15=25</math>. As for the line with slope <math>6</math>, in order for it to travel "up" <math>30</math> units it must have traveled <math>30/6=5</math> units to the right. Thus its <math>x</math>-intercept is at <math>x=40-5=35</math>. Then the distance between them is <math>35-25=10 \Rightarrow \boxed{(\text{B}) 10} | In order for the line with slope <math>2</math> to travel "up" <math>30</math> units (from <math>y=0</math>), it must have traveled <math>30/2=15</math> units to the right. Thus, the <math>x</math>-intercept is at <math>x=40-15=25</math>. As for the line with slope <math>6</math>, in order for it to travel "up" <math>30</math> units it must have traveled <math>30/6=5</math> units to the right. Thus its <math>x</math>-intercept is at <math>x=40-5=35</math>. Then the distance between them is <math>35-25=10 \Rightarrow \boxed{(\text{B}) 10} | ||
\indent</math> | \indent</math> |
Revision as of 13:20, 19 January 2021
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?
Solutions
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
Solution 2
In order for the line with slope to travel "up" units (from ), it must have traveled units to the right. Thus, the -intercept is at . As for the line with slope , in order for it to travel "up" units it must have traveled units to the right. Thus its -intercept is at . Then 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 |
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