2018 AMC 12B Problems/Problem 3

Problem

A line with slope 2 intersects a line with slope 6 at the point $(40,30)$ . What is the distance between the $x$ -intercepts of these two lines?

$(\text{A}) 5 \qquad (\text{B}) 10 \qquad (\text{C}) 20 \qquad (\text{D}) 25 \qquad (\text{E}) 50$

Solutions

Solution 1

Using the slope-intercept form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$ . Simplifying, we get $6x-y=210$ and $2x-y=50$ . Letting $y=0$ in both equations and solving for $x$ gives the $x$ -intercepts: $x=35$ and $x=25$ , respectively. Thus the distance between them is $35-25 = 10 \Rightarrow \boxed{(\text{B}) 10} \indent$

Solution 2

In order for the line with slope $2$ to travel "up" $30$ units (from $y=0$ ), it must have traveled $30/2=15$ units to the right. Thus, the $x$ -intercept is at $x=40-15=25$ . As for the line with slope $6$ , in order for it to travel "up" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$ -intercept is at $x=40-5=35$ . Then the distance between them is $35-25=10 \Rightarrow \boxed{(\text{B}) 10} \indent$

2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AMC 12 Problems and Solutions

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2018 AMC 12B Problems/Problem 3

Contents

Problem

Solutions

Solution 1

Solution 2

See Also