AMC 10 · 2003 · #11
Grade 8 algebraA line with slope 3 intersects a line with slope 5 at point (10,15). What is the distance between the x-intercepts of these two lines?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two lines both pass through the point $(10,15)$: one has slope $3$ and the other has slope $5$. Each line crosses the $x$-axis at its own $x$-intercept. Find the distance between those two $x$-intercepts.
Givens: One line has slope $3$ and passes through $(10,15)$; The other line has slope $5$ and passes through $(10,15)$; An $x$-intercept is the point where a line meets the $x$-axis, so its $y$-coordinate is $0$; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $12$, (E) $20$
Unknowns: The distance between the two lines' $x$-intercepts
Understand
Restated: Two lines both pass through the point $(10,15)$: one has slope $3$ and the other has slope $5$. Each line crosses the $x$-axis at its own $x$-intercept. Find the distance between those two $x$-intercepts.
Givens: One line has slope $3$ and passes through $(10,15)$; The other line has slope $5$ and passes through $(10,15)$; An $x$-intercept is the point where a line meets the $x$-axis, so its $y$-coordinate is $0$; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $12$, (E) $20$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #8 Analyze the Units, #1 Draw a Diagram
Each line is fully determined by its slope and the shared point $(10,15)$, so tool #4 (Introduce a Variable) says: write the point-slope equation of each line with $x$ and $y$ as the variables, then set $y=0$ to solve for the unknown $x$-intercept. Tool #8 (Analyze the Units) gives the same result faster: slope is rise over run, so to drop the $15$ units from $y=15$ down to $y=0$ each line moves a run of $15/\text{slope}$ backward, telling you exactly how far the intercept sits from $x=10$. Tool #1 (Draw a Diagram) keeps the picture straight — two lines fanning out from one point down to two nearby spots on the $x$-axis — so the final subtraction of the two $x$-values is clearly the distance asked for.
Execute — Answer: A
8.F.B.4 Step 1 Write each line's equation
- A line's point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1,y_1)$ is a known point on it.
- Both lines pass through $(10,15)$, so plug that in as $(x_1,y_1)$.
- The slope-$3$ line is $y - 15 = 3(x - 10)$ and the slope-$5$ line is $y - 15 = 5(x - 10)$.
💡 A slope plus one point it passes through is all you need to write a line's exact equation.
8.EE.C.7 Step 2 Find the slope-3 x-intercept
- An $x$-intercept sits on the $x$-axis, where $y = 0$.
- Put $y = 0$ into the slope-$3$ equation: $0 - 15 = 3(x - 10)$, so $-15 = 3(x - 10)$.
- Divide both sides by $3$ to get $x - 10 = -5$, hence $x = 5$.
- The slope-$3$ line crosses the $x$-axis at $x = 5$.
💡 Setting $y=0$ turns the line's equation into a single equation for the exact spot where it meets the $x$-axis.
8.EE.C.7 Step 3 Find the slope-5 x-intercept
- Do the same with the slope-$5$ line: set $y = 0$ to get $-15 = 5(x - 10)$.
- Divide both sides by $5$ to get $x - 10 = -3$, hence $x = 7$.
- The slope-$5$ line crosses the $x$-axis at $x = 7$.
- A steeper line reaches the axis closer to $x=10$, which fits: $7$ is nearer to $10$ than $5$ is.
💡 The steeper the slope, the shorter the run needed to drop the same $15$ units, so its intercept lands nearer to $x=10$.
6.NS.C.7 Step 4 Subtract to get the distance
- Both intercepts, $x = 5$ and $x = 7$, lie on the $x$-axis, so the distance between them is the gap between their $x$-coordinates: $|7 - 5| = 2$.
- So the distance between the two $x$-intercepts is $2$, which is choice (A).
💡 Two points on the same horizontal line are apart by the absolute difference of their $x$-values.
8.F.B.4 A line's point-slope form is $y - y_1 = m(x - x_1)$, where $m$ is the slope and 8.EE.C.7 An $x$-intercept sits on the $x$-axis, where $y = 0$. Put $y = 0$ into the slope 8.EE.C.7 Do the same with the slope-$5$ line: set $y = 0$ to get $-15 = 5(x - 10)$. Divid 6.NS.C.7 Both intercepts, $x = 5$ and $x = 7$, lie on the $x$-axis, so the distance betwe Review
Reasonableness: Both intercepts, $5$ and $7$, sit just to the left of $x=10$, which makes sense because both lines slope upward and must come back down to the axis. The steeper slope-$5$ line lands at $7$ (closer to $10$) and the gentler slope-$3$ line at $5$ (farther), a gap of $2$ — small, and matching choice (A). Checking a point: on the slope-$3$ line at $x=5$, $y = 15 + 3(5-10) = 15 - 15 = 0$, and on the slope-$5$ line at $x=7$, $y = 15 + 5(7-10) = 15 - 15 = 0$, so both really do hit the axis.
Alternative: Skip the full equations. Each line drops the same $15$ units from $(10,15)$ down to the $x$-axis, and slope is rise over run, so the run is $15/\text{slope}$. That places the intercepts at $10 - \frac{15}{3} = 5$ and $10 - \frac{15}{5} = 7$. Since both share the same starting $x=10$, the distance is just $\frac{15}{3} - \frac{15}{5} = 5 - 3 = 2$, again (A).
CCSS standards used (min grade 8)
8.F.B.4Construct a function to model a linear relationship between two quantities (Writing each line's point-slope equation from its slope and the shared point $(10,15)$.)8.EE.C.7Solve linear equations in one variable (Setting $y=0$ and solving $-15 = m(x-10)$ for each line's $x$-intercept.)6.NS.C.7Understand ordering and absolute value of rational numbers (Taking $|7-5|$ as the distance between the two intercepts on the $x$-axis.)
⭐ To find where a line hits the $x$-axis, set $y=0$ and solve for $x$; the distance between two intercepts is just the difference of their $x$-values.
⭐ To find where a line hits the $x$-axis, set $y=0$ and solve for $x$; the distance between two intercepts is just the difference of their $x$-values.
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