Gradient of an Interval
The gradient of an interval measures its steepness as rise over run, \(m=\dfrac{y_2-y_1}{x_2-x_1}\). A positive gradient slopes up to the right and a negative gradient slopes down. In NSW Year 10 Advanced maths you find gradients from coordinates that include fractions, decimals and surds, and use them to find unknowns (NESA outcome MA5‑LIN‑P‑01).
You will find gradients of intervals, solve for a missing coordinate when the gradient is known, and use gradients to show that three points lie on one line or that a quadrilateral is a trapezium. Real-world questions find the gradient of a ski run and a railway line as a percentage grade, and the rate at which a hot-air balloon comes down. Harder questions use points with pronumeral coordinates and prove a general result about gradients. It is part of Year 10 Advanced Straight Line Graphs.
Every question with a fully worked solution.
- Gradient of an Interval - Video - Gradient Formula Watch
Theory
The gradient of an interval measures its steepness: \(m = \dfrac{\text{rise}}{\text{run}} = \dfrac{y_2 - y_1}{x_2 - x_1}\). This Year 10 Advanced maths page covers the gradient formula with fractions and surds, finding unknown coordinates, collinear points and parallel sides, and gradient as a rate of change.
The gradient \(m\) of an interval is how far it rises for each unit it runs across. The rise is the vertical change, \(y_2 - y_1\), and the run is the horizontal change, \(x_2 - x_1\).
Reading from left to right, a line going up has a positive gradient and a line going down has a negative gradient. A horizontal line has gradient \(0\). A vertical line has no run, so its gradient is undefined.
The bigger the size of the gradient, the steeper the line. Steepness ignores the sign: gradients \(3\) and \(-3\) are equally steep.
Parallel lines have equal gradients. Points are collinear if they lie on one straight line, so every pair of them gives the same gradient. On a graph against time, the gradient is a rate of change, such as litres per minute.
The gradient of the interval joining \((x_1,\ y_1)\) and \((x_2,\ y_2)\):
How to find a gradient
- Label the two points. Either can be first, but use the same order on the top and the bottom.
- Subtract the \(y\)-values for the rise and the \(x\)-values for the run, with brackets round negatives.
- Divide and simplify. Check the sign against a sketch: uphill positive, downhill negative.
Using gradients to solve problems
- Unknown coordinate: put the pronumeral into the formula, set it equal to the given gradient and solve (Example 2). If the point lies on a line such as \(y = 2x - 1\), write it as \((t,\ 2t - 1)\) and cross-multiply.
- Collinear points: \(m_{AB} = m_{BC}\). If the unknown appears squared, you get a quadratic; factorise and keep every solution that works.
- Linked gradients: if one gradient is twice, or \(1\) less than, another, find the known one first, then write and solve the equation.
- Classifying shapes: equal gradients mean parallel sides. Two pairs give a parallelogram, one pair a trapezium (Example 3).
- Rates of change: the gradient of a graph against time is the rate, with units such as m per minute. Use it to find when a value is reached (Example 4).
- Steepness comparisons: compare sizes of gradients. If a downhill section is involved, its steepness is the gradient without the negative sign.
- General results: with pronumerals in both points, write the top as a single fraction and look for a factor that matches the bottom, such as \(a - b = -(b - a)\).
(i) \((-2,\ 5)\) and \((4,\ -4)\)
(ii) \((\sqrt{3},\ 2)\) and \((\sqrt{12},\ 8)\), in simplest surd form.
(i) Subtract in the same order on the top and the bottom.
| \(m\) | \(=\) | \(\dfrac{y_2 - y_1}{x_2 - x_1}\) |
| \(\) | \(=\) | \(\dfrac{-4 - 5}{4 - (-2)}\) |
| \(\) | \(=\) | \(\dfrac{-9}{6} = \mathbf{-\dfrac{3}{2}}\) |
The gradient is negative, so the line falls from left to right.
(ii) Simplify \(\sqrt{12} = 2\sqrt{3}\) first, then rationalise the denominator.
| \(m\) | \(=\) | \(\dfrac{8 - 2}{2\sqrt{3} - \sqrt{3}} = \dfrac{6}{\sqrt{3}}\) |
| \(\) | \(=\) | \(\dfrac{6}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{6\sqrt{3}}{3}\) |
| \(\) | \(=\) | \(\mathbf{2\sqrt{3}}\) |
Substitute into the gradient formula with the unknown in place, then solve the equation.
| \(\dfrac{7 - k}{-4 - 2}\) | \(=\) | \(-\dfrac{2}{3}\) |
| \(\dfrac{7 - k}{-6}\) | \(=\) | \(-\dfrac{2}{3}\) |
| \(7 - k\) | \(=\) | \(-\dfrac{2}{3} \times (-6)\) |
| \(7 - k\) | \(=\) | \(4\) |
| \(k\) | \(=\) | \(\mathbf{3}\) |
Check: \(\dfrac{7 - 3}{-4 - 2} = \dfrac{4}{-6} = -\dfrac{2}{3}\ \checkmark\)
Parallel lines have equal gradients. Find the gradient of each side.
| \(m_{AB}\) | \(=\) | \(\dfrac{1 - (-2)}{3 - (-3)} = \dfrac{3}{6} = \dfrac{1}{2}\) |
| \(m_{DC}\) | \(=\) | \(\dfrac{4 - 1}{4 - (-2)} = \dfrac{3}{6} = \dfrac{1}{2}\) |
| \(m_{BC}\) | \(=\) | \(\dfrac{4 - 1}{4 - 3} = 3\) |
| \(m_{AD}\) | \(=\) | \(\dfrac{1 - (-2)}{-2 - (-3)} = 3\) |
\(AB \parallel DC\) and \(BC \parallel AD\). Both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram. With only one pair parallel it would be a trapezium.
(i) Find the gradient of the graph of volume against time, in litres per minute, and explain its meaning.
(ii) At what time will the tank be empty?
(i) Time is on the horizontal axis. From 2:15 to 2:45 is \(30\) minutes.
| \(m\) | \(=\) | \(\dfrac{2700 - 4200}{30}\) |
| \(\) | \(=\) | \(\dfrac{-1500}{30} = \mathbf{-50}\) |
The tank loses \(50\) litres every minute; the negative sign means the volume is falling.
(ii) At 2:45 pm \(2700\) L remain.
| \(\text{Time to empty}\) | \(=\) | \(\dfrac{2700}{50} = 54\text{ minutes}\) |
\(54\) minutes after 2:45 pm is 3:39 pm.
Common pitfalls
Frequently asked questions
What is the gradient formula?
The gradient of the line through \((x_1,\ y_1)\) and \((x_2,\ y_2)\) is \(m = \dfrac{y_2 - y_1}{x_2 - x_1}\). It is the rise divided by the run.
How do you know 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. A flat line has gradient zero.
How do you find a missing coordinate given the gradient?
Substitute both points into the gradient formula, leaving the unknown as a pronumeral. Set the result equal to the given gradient and solve the equation.
How do you show three points are collinear?
Find the gradient from the first point to the second and from the second to the third. If the gradients are equal, the points lie on one straight line.
What does the gradient of a graph mean in real life?
It is a rate of change: how much the vertical quantity changes for each unit of the horizontal quantity. On a distance against time graph it is the speed.