The Gradient Of A Straight Line
The gradient of a straight line is a foundational skill assumed for Queensland Year 11 Mathematical Methods (QCAA). The gradient measures a line's steepness and direction, found from two points as the rise over the run.
You will calculate a gradient from two points, read its sign and size, handle horizontal and vertical lines, and find an unknown coordinate from a given gradient — groundwork for the equation of a line.
Every question with a fully worked solution.
- The Gradient Of A Straight Line - Video - Gradient of a straight line through two points Watch
Theory
In Year 11 Mathematical Methods (QCAA), the gradient of a straight line measures its steepness and direction. From two points it is the rise over the run, \(m=\dfrac{y_2-y_1}{x_2-x_1}\). This page shows how to compute a gradient from two points, read the sign and size, handle horizontal and vertical lines, and find an unknown coordinate from a given gradient.
The gradient \(m\) of a straight line is the amount \(y\) changes for each unit that \(x\) increases. Between two points it is the rise (the vertical change) divided by the run (the horizontal change).
The sign tells you the direction: a positive gradient rises to the right (increasing), a negative gradient falls to the right (decreasing). The size tells you the steepness — a larger absolute value is steeper. A horizontal line has gradient \(0\); a vertical line has an undefined gradient.
The gradient through \(A(x_1,y_1)\) and \(B(x_2,y_2)\) is
How to find a gradient from two points
- Label the points \((x_1,y_1)\) and \((x_2,y_2)\).
- Rise: subtract the \(y\)-values, \(y_2-y_1\). Run: subtract the \(x\)-values in the same order, \(x_2-x_1\).
- Divide rise by run and simplify; then read off the sign (increasing or decreasing) and the size (steepness).
Rise — subtract the \(y\)-values:
| \(y_2-y_1\) | \(=\) | \(11-2\) |
| \(=\) | \(9\) |
Run — subtract the \(x\)-values in the same order:
| \(x_2-x_1\) | \(=\) | \(4-1\) |
| \(=\) | \(3\) |
Divide rise by run:
| \(m\) | \(=\) | \(\dfrac{9}{3}\) |
| \(=\) | \(3\) |
Gradient: \(m=3\) (the line increases).
Rise — subtract the \(y\)-values:
| \(y_2-y_1\) | \(=\) | \(-7-5\) |
| \(=\) | \(-12\) |
Run — subtract the \(x\)-values in the same order:
| \(x_2-x_1\) | \(=\) | \(4-(-2)\) |
| \(=\) | \(6\) |
Divide rise by run:
| \(m\) | \(=\) | \(\dfrac{-12}{6}\) |
| \(=\) | \(-2\) |
Gradient: \(m=-2\) (the line decreases).
Line (a) — the \(y\)-values are equal, so the rise is \(0\):
| \(m\) | \(=\) | \(\dfrac{3-3}{7-2}\) |
| \(=\) | \(\dfrac{0}{5}\) | |
| \(=\) | \(0\) |
Line (a) is horizontal, so its gradient is \(0\).
Line (b) — the \(x\)-values are equal, so the run is \(0\):
| \(m\) | \(=\) | \(\dfrac{6-1}{4-4}\) |
| \(=\) | \(\dfrac{5}{0}\) |
Dividing by \(0\) is not defined, so line (b) is vertical with an undefined gradient.
(a) \(m=0\); (b) gradient undefined (vertical line).
Write the gradient formula and substitute:
| \(m\) | \(=\) | \(\dfrac{1-k}{6-2}\) |
| \(3\) | \(=\) | \(\dfrac{1-k}{4}\) |
Multiply both sides by \(4\), then solve:
| \(3\times 4\) | \(=\) | \(1-k\) |
| \(12\) | \(=\) | \(1-k\) |
| \(k\) | \(=\) | \(1-12\) |
| \(k\) | \(=\) | \(-11\) |
Unknown coordinate: \(k=-11\).
Common pitfalls
Frequently asked questions
How do you find the gradient between two points?
Subtract the \(y\)-values for the rise and the \(x\)-values for the run in the same order, then divide: \(m=\dfrac{y_2-y_1}{x_2-x_1}\).
What does a negative gradient mean?
The line falls from left to right — as \(x\) increases, \(y\) decreases. A positive gradient rises to the right.
What is the gradient of a horizontal line?
Zero. The \(y\)-values are equal, so the rise is \(0\) and \(m=\dfrac{0}{\text{run}}=0\).
Why is a vertical line's gradient undefined?
Both points have the same \(x\)-value, so the run is \(0\). Dividing by \(0\) is not defined, so the gradient does not exist.
Which gradient is steeper, 2 or minus 3?
The one with the larger size (absolute value). \(|-3|=3 > 2\), so a gradient of \(-3\) is steeper than \(2\).