Gradient of an Interval
Gradient of an Interval in NSW Year 10 Core maths measures how steep a line is (NESA outcome MA5‑LIN‑C‑01). The gradient is rise divided by run: positive lines go up to the right, negative lines go down, horizontal lines have gradient 0 and vertical lines have an undefined gradient.
You will find gradients from diagrams, from two points and from a table, give them as fractions or decimals, and work backwards to a missing coordinate. You will also find the slope of a roof, the rise of a road from its grade, check whether a driveway meets council rules, and test whether three points lie on one straight line. It is part of Year 10 Core Straight Line Graphs.
Every question with a fully worked solution.
- Gradient of an Interval - Video - Gradient Formula Watch
Theory
This Year 10 maths page shows how to find the gradient of an interval: rise over run from a graph or a table, the gradient formula for two points, what a zero or undefined gradient means, how steep a roof, road or ramp is, and how to use the gradient to find a missing coordinate.
The gradient (slope) \(m\) of an interval measures how steep it is. It is the rise (change in \(y\)) divided by the run (change in \(x\)).
Read an interval from left to right. If it goes up, the gradient is positive. If it goes down, the rise is negative, so the gradient is negative.
A horizontal interval has a rise of \(0\), so its gradient is \(0\). A vertical interval has a run of \(0\). Dividing by \(0\) is not possible, so its gradient is undefined.
The bigger the number (ignoring its sign), the steeper the interval: a gradient of \(-3\) is steeper than a gradient of \(2\).
Points that lie on one straight line are called collinear. Any two of them give the same gradient.
Gradient as rise over run, and the gradient formula for the points \((x_1, y_1)\) and \((x_2, y_2)\):
How to find the gradient of an interval
- From a graph: draw the right-angled triangle. Count the run to the right, then the rise (up is \(+\), down is \(-\)).
- From two points: label them \((x_1, y_1)\) and \((x_2, y_2)\). Either point can be first, but keep the same order on the top and the bottom.
- Substitute into \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\), with negatives in brackets.
- Simplify the fraction. Give a decimal if the question asks.
- Check the sign against a quick sketch: up to the right is positive.
How to find an unknown coordinate
- Substitute both points into the gradient formula and make it equal to the given gradient.
- Solve the equation (cross-multiply if there is a fraction on each side), then check.
How to test for points on one line
- Find the gradient of \(AB\) and of \(BC\).
- Compare: equal gradients through the shared point \(B\) mean one line. Different gradients mean the points are not collinear.
(i) \((-1, -4)\) and \((3, 6)\)
(ii) \((-2, 5)\) and \((6, 5)\)
(i) Label the points: \((x_1, y_1) = (-1, -4)\) and \((x_2, y_2) = (3, 6)\). Put negatives in brackets.
| \(m\) | \(=\) | \(\dfrac{y_2 - y_1}{x_2 - x_1}\) |
| \(\) | \(=\) | \(\dfrac{6 - (-4)}{3 - (-1)}\) |
| \(\) | \(=\) | \(\dfrac{10}{4}\) |
| \(\) | \(=\) | \(\boldsymbol{\dfrac{5}{2}}\) |
(ii) The \(y\)-values are the same, so the rise is \(0\).
| \(m\) | \(=\) | \(\dfrac{5 - 5}{6 - (-2)} = \dfrac{0}{8}\) |
| \(\) | \(=\) | \(\boldsymbol{0}\) |
The interval is horizontal.
(i) Find its gradient as a fraction and as a percentage grade.
(ii) A plan says a ramp can be no steeper than \(1\) in \(14\). Does this ramp meet the plan?
(i) Use the same units: \(45~\text{cm} = 0.45~\text{m}\).
| \(m\) | \(=\) | \(\dfrac{\text{rise}}{\text{run}} = \dfrac{0.45}{6}\) |
| \(\) | \(=\) | \(\dfrac{3}{40} = 0.075\) |
| \(\text{grade}\) | \(=\) | \(\boldsymbol{0.075 \times 100\% = 7.5\%}\) |
(ii) \(1\) in \(14\) means a gradient of \(\dfrac{1}{14} \approx 0.0714\).
| \(0.075\) | \(>\) | \(0.0714\) |
The ramp is steeper than the limit, so it does not meet the plan.
Put the points into the gradient formula and make it equal to the gradient.
| \(\dfrac{-2 - 4}{t - (-1)}\) | \(=\) | \(-\dfrac{3}{2}\) |
| \(\dfrac{-6}{t + 1}\) | \(=\) | \(-\dfrac{3}{2}\) |
Cross-multiply:
| \(-6 \times 2\) | \(=\) | \(-3(t + 1)\) |
| \(-12\) | \(=\) | \(-3t - 3\) |
| \(-9\) | \(=\) | \(-3t\) |
| \(t\) | \(=\) | \(\boldsymbol{3}\) |
Check: \(\dfrac{-2 - 4}{3 + 1} = \dfrac{-6}{4} = -\dfrac{3}{2}\) ✓
(i) Show that \(A(-3, -1)\), \(B(1, 1)\) and \(C(9, 5)\) lie on one straight line.
(ii) The point \(D(13, k)\) is on the same line. Find \(k\).
(i) Points are on one line when the gradient between any two of them is the same.
| \(m_{AB}\) | \(=\) | \(\dfrac{1 - (-1)}{1 - (-3)} = \dfrac{2}{4} = \dfrac{1}{2}\) |
| \(m_{BC}\) | \(=\) | \(\dfrac{5 - 1}{9 - 1} = \dfrac{4}{8} = \dfrac{1}{2}\) |
The gradients are equal and the intervals share the point \(B\), so \(A\), \(B\) and \(C\) are on one line.
(ii) The gradient of \(CD\) must also be \(\dfrac{1}{2}\).
| \(\dfrac{k - 5}{13 - 9}\) | \(=\) | \(\dfrac{1}{2}\) |
| \(k - 5\) | \(=\) | \(2\) |
| \(k\) | \(=\) | \(\boldsymbol{7}\) |
Common pitfalls
Frequently asked questions
What is the gradient formula?
For the points \((x_1, y_1)\) and \((x_2, y_2)\), \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). It is rise over run written with coordinates.
How can you tell if a gradient is positive or negative?
Read the line from left to right. If it goes up, the gradient is positive. If it goes down, the gradient is negative.
Why is the gradient of a vertical line undefined?
A vertical line has a run of \(0\). The gradient would mean dividing by \(0\), which cannot be done, so the gradient is undefined.
What does a road grade of 6 percent mean?
It means a gradient of \(0.06\): the road rises \(6~\text{m}\) for every \(100~\text{m}\) measured horizontally.
How do you know if three points are on one straight line?
Find the gradient from the first point to the second and from the second to the third. If the two gradients are equal, the points are on one line.