Topics

The gradient of a line

The formula

The gradient measures how steep a line is and which way it slopes:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Often remembered as $\dfrac{\text{rise}}{\text{run}}$ — how far up for how far along.
Each part:
- $(y_2 - y_1)$ — the rise, the vertical change
- $(x_2 - x_1)$ — the run, the horizontal change
- $m$ — the gradient, which may be positive, negative, zero or undefined

Keeping the order consistent

Unlike the distance formula, the order matters here — but only if you are inconsistent.
Both differences must be taken in the same direction. Subtracting $y$ one way and $x$ the other flips the sign.

$\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 - y_2}{x_1 - x_2} \qquad \text{but} \qquad \frac{y_2 - y_1}{x_1 - x_2} \; \text{is wrong}$

Write the two points down and label them before substituting.

What the value means

Positive gradient — the line rises from left to right
Negative gradient — the line falls from left to right
Zero gradient — the line is horizontal; $y_2 = y_1$ so the rise is 0
Undefined gradient — the line is vertical; $x_2 = x_1$ so the run is 0 and you cannot divide by zero
A steeper line has a larger gradient in magnitude. A gradient of $-5$ is steeper than one of $2$ , even though it is negative.
Do not say a vertical line has "zero gradient" or "infinite gradient". The correct statement is that its gradient is undefined.

Working through an example

The gradient of the line through $A(2, 5)$ and $B(8, 17)$ :

$m = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2$

With negatives, for $C(-3, 4)$ and $D(5, -8)$ :

$m = \frac{-8 - 4}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}$

Simplify the fraction, and leave it as a fraction rather than a decimal where it does not terminate neatly.

Collinear points

Three or more points are collinear if they all lie on the same straight line.
The test: find the gradient between the first pair and between the second pair. If the gradients are equal, the points are collinear.
They must share a point. Compare $AB$ with $BC$ , so that $B$ is common to both — equal gradients then force all three onto one line.
Worked through — are $P(1, 2)$ , $Q(4, 8)$ and $R(6, 12)$ collinear?
- $m_{PQ} = \dfrac{8-2}{4-1} = \dfrac{6}{3} = 2$
- $m_{QR} = \dfrac{12-8}{6-4} = \dfrac{4}{2} = 2$
- Equal gradients through the shared point $Q$ , so $P$ , $Q$ and $R$ are collinear ✓

Gradient as a rate of change

In a context, the gradient is a rate, and its units come from the axes.
If the horizontal axis is time in hours and the vertical axis is distance in km, the gradient is in km per hour — a speed.
Interpreting the gradient in context is a Merit-level skill: not just "the gradient is 4", but "the tank is filling at 4 litres per minute".

Tips and Tricks

Keep the subtraction consistent. Both the top and the bottom must run from the same starting point. Label the points $(x_1, y_1)$ and $(x_2, y_2)$ explicitly before you substitute, and the sign takes care of itself.

$y$ on top, $x$ on the bottom. Inverting the fraction is a frequent slip and gives the reciprocal of the right answer — which looks plausible enough to survive unnoticed.

A vertical line has an UNDEFINED gradient, not zero. Zero belongs to horizontal lines. Getting these the wrong way round is a classic error.

Simplify the fraction and keep it as a fraction. $\frac{6}{4}$ should be written $\frac{3}{2}$ , and leaving it exact avoids rounding in later steps.

For collinearity, use a shared point. Comparing $m_{AB}$ with $m_{CD}$ shows only that two lines are parallel; comparing $m_{AB}$ with $m_{BC}$ proves the three points lie on one line.

In a context, say what the gradient means with its units. That interpretation is the difference between an Achieved answer and a Merit one.

Worked ExampleTesting for collinearity and interpreting a gradient

(a) Determine whether $A(-2, 9)$ , $B(1, 3)$ and $C(4, -3)$ are collinear. (b) A tank's volume $V$ litres at time $t$ minutes passes through $(2, 45)$ and $(10, 13)$ . Find the gradient and interpret it.

Part (a), Step 1 — Choose which gradients to compare

To test collinearity, compare gradients that share a point. Using $B$ as the shared point, we compare $m_{AB}$ and $m_{BC}$ .

Part (a), Step 2 — Find $m_{AB}$

Taking $A(-2, 9)$ as point 1 and $B(1, 3)$ as point 2:

$m_{AB} = \frac{3 - 9}{1 - (-2)} = \frac{-6}{3} = -2$

Note the bracket: $1 - (-2) = 3$ , not $-1$ .

Part (a), Step 3 — Find $m_{BC}$

Taking $B(1, 3)$ as point 1 and $C(4, -3)$ as point 2 — the same direction of subtraction:

$m_{BC} = \frac{-3 - 3}{4 - 1} = \frac{-6}{3} = -2$

Part (a), Step 4 — Compare and conclude

$m_{AB} = m_{BC} = -2$

The two gradients are equal, and the segments share the point $B$ . Two segments with the same gradient through a common point must lie along the same straight line.

Part (b), Step 1 — Identify the variables and their units

Horizontal axis: time $t$ , in minutes
Vertical axis: volume $V$ , in litres

So the gradient will be in litres per minute.

Part (b), Step 2 — Calculate the gradient

$m = \frac{13 - 45}{10 - 2} = \frac{-32}{8} = -4$

Part (b), Step 3 — Interpret in context

The gradient is $-4$ litres per minute.

The magnitude, 4, says the volume changes by 4 litres every minute.
The negative sign says the volume is decreasing.

Part (b), Step 4 — Extend the interpretation

Because the gradient is constant, the rate is steady — the tank drains at the same speed throughout, which is what a straight-line model assumes.

Working backwards, at $t = 2$ the volume was 45 L, so at $t = 0$ it would have been $45 + 2(4) = 53$ L, and the tank would be empty when

$53 - 4t = 0 \quad \Rightarrow \quad t = 13.25 \text{ minutes}$

The model only applies while there is water in the tank — beyond $t = 13.25$ minutes it would predict a negative volume, which is impossible.

The gradient of a line

Covers AS91256

The formula

The gradient measures how steep a line is and which way it slopes:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

Often remembered as $\dfrac{\text{rise}}{\text{run}}$ — how far up for how far along.
Each part:
- $(y_2 - y_1)$ — the rise, the vertical change
- $(x_2 - x_1)$ — the run, the horizontal change
- $m$ — the gradient, which may be positive, negative, zero or undefined

Keeping the order consistent

Unlike the distance formula, the order matters here — but only if you are inconsistent.
Both differences must be taken in the same direction. Subtracting $y$ one way and $x$ the other flips the sign.

$\frac{y_2 - y_1}{x_2 - x_1} = \frac{y_1 - y_2}{x_1 - x_2} \qquad \text{but} \qquad \frac{y_2 - y_1}{x_1 - x_2} \; \text{is wrong}$

Write the two points down and label them before substituting.

What the value means

Positive gradient — the line rises from left to right
Negative gradient — the line falls from left to right
Zero gradient — the line is horizontal; $y_2 = y_1$ so the rise is 0
Undefined gradient — the line is vertical; $x_2 = x_1$ so the run is 0 and you cannot divide by zero
A steeper line has a larger gradient in magnitude. A gradient of $-5$ is steeper than one of $2$ , even though it is negative.
Do not say a vertical line has "zero gradient" or "infinite gradient". The correct statement is that its gradient is undefined.

Working through an example

The gradient of the line through $A(2, 5)$ and $B(8, 17)$ :

$m = \frac{17 - 5}{8 - 2} = \frac{12}{6} = 2$

With negatives, for $C(-3, 4)$ and $D(5, -8)$ :

$m = \frac{-8 - 4}{5 - (-3)} = \frac{-12}{8} = -\frac{3}{2}$

Simplify the fraction, and leave it as a fraction rather than a decimal where it does not terminate neatly.

Collinear points

Three or more points are collinear if they all lie on the same straight line.
The test: find the gradient between the first pair and between the second pair. If the gradients are equal, the points are collinear.
They must share a point. Compare $AB$ with $BC$ , so that $B$ is common to both — equal gradients then force all three onto one line.
Worked through — are $P(1, 2)$ , $Q(4, 8)$ and $R(6, 12)$ collinear?
- $m_{PQ} = \dfrac{8-2}{4-1} = \dfrac{6}{3} = 2$
- $m_{QR} = \dfrac{12-8}{6-4} = \dfrac{4}{2} = 2$
- Equal gradients through the shared point $Q$ , so $P$ , $Q$ and $R$ are collinear ✓

Gradient as a rate of change

In a context, the gradient is a rate, and its units come from the axes.
If the horizontal axis is time in hours and the vertical axis is distance in km, the gradient is in km per hour — a speed.
Interpreting the gradient in context is a Merit-level skill: not just "the gradient is 4", but "the tank is filling at 4 litres per minute".

Tips and Tricks

$y$ on top, $x$ on the bottom. Inverting the fraction is a frequent slip and gives the reciprocal of the right answer — which looks plausible enough to survive unnoticed.

A vertical line has an UNDEFINED gradient, not zero. Zero belongs to horizontal lines. Getting these the wrong way round is a classic error.

Simplify the fraction and keep it as a fraction. $\frac{6}{4}$ should be written $\frac{3}{2}$ , and leaving it exact avoids rounding in later steps.

For collinearity, use a shared point. Comparing $m_{AB}$ with $m_{CD}$ shows only that two lines are parallel; comparing $m_{AB}$ with $m_{BC}$ proves the three points lie on one line.

In a context, say what the gradient means with its units. That interpretation is the difference between an Achieved answer and a Merit one.

Worked ExampleTesting for collinearity and interpreting a gradient

Part (a), Step 1 — Choose which gradients to compare

To test collinearity, compare gradients that share a point. Using $B$ as the shared point, we compare $m_{AB}$ and $m_{BC}$ .

Part (a), Step 2 — Find $m_{AB}$

Taking $A(-2, 9)$ as point 1 and $B(1, 3)$ as point 2:

$m_{AB} = \frac{3 - 9}{1 - (-2)} = \frac{-6}{3} = -2$

Note the bracket: $1 - (-2) = 3$ , not $-1$ .

Part (a), Step 3 — Find $m_{BC}$

Taking $B(1, 3)$ as point 1 and $C(4, -3)$ as point 2 — the same direction of subtraction:

$m_{BC} = \frac{-3 - 3}{4 - 1} = \frac{-6}{3} = -2$

Part (a), Step 4 — Compare and conclude

$m_{AB} = m_{BC} = -2$

The two gradients are equal, and the segments share the point $B$ . Two segments with the same gradient through a common point must lie along the same straight line.

Part (b), Step 1 — Identify the variables and their units

Horizontal axis: time $t$ , in minutes
Vertical axis: volume $V$ , in litres

So the gradient will be in litres per minute.

Part (b), Step 2 — Calculate the gradient

$m = \frac{13 - 45}{10 - 2} = \frac{-32}{8} = -4$

Part (b), Step 3 — Interpret in context

The gradient is $-4$ litres per minute.

The magnitude, 4, says the volume changes by 4 litres every minute.
The negative sign says the volume is decreasing.

Part (b), Step 4 — Extend the interpretation

Because the gradient is constant, the rate is steady — the tank drains at the same speed throughout, which is what a straight-line model assumes.

Working backwards, at $t = 2$ the volume was 45 L, so at $t = 0$ it would have been $45 + 2(4) = 53$ L, and the tank would be empty when

$53 - 4t = 0 \quad \Rightarrow \quad t = 13.25 \text{ minutes}$

The model only applies while there is water in the tank — beyond $t = 13.25$ minutes it would predict a negative volume, which is impossible.

The gradient of a line

The formula

Keeping the order consistent

What the value means

Working through an example

Collinear points

Gradient as a rate of change

Part (a), Step 1 — Choose which gradients to compare

Part (a), Step 2 — Find mABm_{AB}mAB​

Part (a), Step 3 — Find mBCm_{BC}mBC​

Part (a), Step 4 — Compare and conclude

Part (b), Step 1 — Identify the variables and their units

Part (b), Step 2 — Calculate the gradient

Part (b), Step 3 — Interpret in context

Part (b), Step 4 — Extend the interpretation

The gradient of a line

The formula

Keeping the order consistent

What the value means

Working through an example

Collinear points

Gradient as a rate of change

Part (a), Step 1 — Choose which gradients to compare

Part (a), Step 2 — Find mABm_{AB}mAB​

Part (a), Step 3 — Find mBCm_{BC}mBC​

Part (a), Step 4 — Compare and conclude

Part (b), Step 1 — Identify the variables and their units

Part (b), Step 2 — Calculate the gradient

Part (b), Step 3 — Interpret in context

Part (b), Step 4 — Extend the interpretation

Part (a), Step 2 — Find $m_{AB}$

Part (a), Step 3 — Find $m_{BC}$

Part (a), Step 2 — Find $m_{AB}$

Part (a), Step 3 — Find $m_{BC}$