2003 AMC 10B Problem 11

Both lines pass through $(10, 15)$ and must descend $15$ units to reach the $x$-axis.

On the line of slope $3$, each unit of horizontal movement changes $y$ by $3$, so descending $15$ requires moving $\frac{15}{3} = 5$ units to the left. Its $x$-intercept is $10 - 5 = 5$.

On the line of slope $5$, descending $15$ requires $\frac{15}{5} = 3$ units to the left, so its $x$-intercept is $10 - 3 = 7$.

The distance between the intercepts is $7 - 5 = 2$.

(Equivalently, the lines are $y = 3x - 15$ and $y = 5x - 35$; setting $y = 0$ gives $x = 5$ and $x = 7$.)

The answer is $\boxed{\textbf{(A)}\ 2}$.

Slope is rise over run, so a known vertical drop translates directly into a horizontal run of $\frac{\text{drop}}{\text{slope}}$. Thinking of the intercept as "how far left do I travel to fall $15$" avoids writing equations at all and keeps the arithmetic to two divisions. Steeper lines reach the axis sooner, which also explains why the slope-$5$ intercept is closer to $x = 10$.

• Finding the $y$-intercepts ($-15$ and $-35$) instead of the $x$-intercepts, giving a difference of $20$.
• Sign slip when moving left from $x = 10$, e.g. reporting intercepts $15$ and $13$; the difference happens to still be $2$, but the method would fail on similar problems.