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?

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


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 $35-25=10 \Rightarrow \boxed{(\text{B}) 10} \indent$

See Also

2018 AMC 12B (ProblemsAnswer KeyResources)
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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