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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

Learn8300Papers 1–3 · Higher tierAlgebraGradients and straight lines

Algebra · Subtopic 8 of 11

Gradients and Straight Lines

Two points fix a straight line. From them you can find how steep it is (gradient), how long the segment is (Pythagoras), its middle (average the coordinates), and its equation $y=mx+c$ .

Cambridge 4024 Paper 1Cambridge 4024 Paper 2Edexcel 4MA1 Papers 1H, 2HEdexcel 4MA1 Papers 1F, 2FAQA 8300 Papers 1–3≈ 30 minRevision 6 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Find the gradient of a line through two points.
Find the length and midpoint of a line segment.
Find the equation of a line in the form $y=mx+c$ .
Read the gradient and intercept from an equation such as $2x+3y=12$ .

Pause and think

A road sign warns of a " $10\%$ " hill. What do you think $10\%$ measures?

Section 2 of 14

Core idea

Start hereHow steep is the road?

A road sign says "1 in 10": rise 1 metre for every 10 along. That's a gradient of 0.1. Every straight line is fixed by its gradient and one point it passes through.

The big idea

$y=mx+c$ : $m$ gradient, $c$ $y$ -intercept. $m=\frac{y_2-y_1}{x_2-x_1}$ .

1. Gradient

y = mx + 1 as the gradient changes

gradient m = -2.0

Positive

m

: uphill. Negative: downhill. Zero: flat. Bigger numbers: steeper.

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

$(x_1,y_1),(x_2,y_2)$: two points on the line.

\text{midpoint}=\left(\tfrac{x_1+x_2}2,\tfrac{y_1+y_2}2\right),\quad\text{length}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

length: from Pythagoras.

2. Finding the equation

Worked exampleThrough $(1,3)$ and $(4,9)$

Equation, midpoint, length.

Goal: Three results.
Strategy: Gradient, then $c$ by substitution.

$m=\frac{9-3}{4-1}=2$
Rise over run.
$3=2(1)+c\Rightarrow c=1$
Substitute a point.
$(2.5,6),\quad\sqrt{9+36}=6.71$
Midpoint, length.

Answer

y=2x+1

;

(2.5,6)

;

6.71

✓ Check

(4,9)

2(4)+1=9

✓.

Worked example $3x+2y=12$

Gradient and intercept.

Goal: $m$ , $c$ .
Strategy: Make $y$ the subject.

$2y=-3x+12$
Subtract $3x$ .
$y=-\tfrac32x+6$
Divide by 2.

Answer

m=-1.5

c=6

✓ Check

(0,6)

0+12=12

✓.

Common misconceptionReading m from 3x + 2y = 12

m

is not 3: rearrange first.

Pause and think

Where does $y=2x-6$ cross the $x$ -axis?

Summary

$y=mx+c$ .
Gradient from two points; substitute for $c$ .

Gradient: Steepness: rise ÷ run.
Intercept: Where a line crosses an axis.

Perpendicular lines: gradients multiply to −1

y=mx+c

m

is the gradient and

c

is where the line crosses the

y

-axis. Here

2\times(-\tfrac12)=-1

, so the lines meet at right angles.

HistoryDescartes and the fly. René Descartes published coordinate geometry in La Géométrie in 1637, linking algebra and shapes. A popular legend says he got the idea lying in bed watching a fly crawl on the ceiling, realising he could describe its position with two distances from the walls.

Fun factThe steepest street. Ffordd Pen Llech in Wales and Baldwin Street in New Zealand have argued over the title of "world's steepest street": the Guinness record moved to Wales in 2019 and back to New Zealand in 2020. Both are around

35\%

, a gradient of about

0.35

: a rise of

3.5

m for every

10

m along.

Section 3 of 14

Key formulae

Gradient

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

Midpoint

\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)

Length

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

Straight line

y=mx+c

TrickKeep the same order. In the gradient formula, if you start with point

B

on top, start with point

B

on the bottom too:

\dfrac{y_B-y_A}{x_B-x_A}

. Mixing the order flips the sign.

Section 4 of 14

Worked examples

Example 1 · Gradient0 / 1 steps

Find the gradient of the line through

(1,2)

and

(4,11)

Example 2 · Equation from point and gradient0 / 2 steps

Find the equation of the line through

(2,7)

with gradient

-1

Example 3 · Reading m and c0 / 2 steps

Find the gradient and

y

-intercept of

2x+3y=12

Section 5 of 14

Explore the graph

The line through two points

Loading the interactive graph…

Try this

Drag the points $(a,b)$ and $(c,d)$ . The line through them and the midpoint update.
Make $d=b$ . What is the gradient of a horizontal line?
Make $c=a$ . Why does the gradient formula break for a vertical line?

Gradient is rise over run; the midpoint averages the coordinates.

In the real worldRamps must not be too steep. Building rules typically limit wheelchair ramps to a gradient of about

1:12

(rising

1

m for every

12

m along). Architects use exactly the gradient formula to check their designs.

Section 6 of 14

Common mistakes

IncorrectGradient

=\dfrac{x_2-x_1}{y_2-y_1}

Correct

\dfrac{y_2-y_1}{x_2-x_1}

Rise (up) over run (across).

Incorrect

2x+3y=12

has gradient

2

Correct

-\tfrac23

Rearrange to

y=\ldots

first.

IncorrectMidpoint by subtracting

CorrectAdd and halve

The midpoint is the average.

IncorrectLength

=(x_2-x_1)+(y_2-y_1)

CorrectUse Pythagoras

The segment is the hypotenuse.

$y=mx+c$

$m$ tells you how steep, $c$ tells you where it crosses. Two numbers, the whole line.

Section 7 of 14

Exam tip

Exam tipLeave answers exact. Lengths often come out as surds:

\sqrt{100}=10

is neat, but

\sqrt{52}

should be left as

\sqrt{52}=2\sqrt{13}

on Paper 1, or given to

3

s.f. on Paper 2. Read the question to see which is wanted.

Section 8 of 14

Guided practice

SA-4024-CG-G014 marksL1 FoundationNon-calculatorGradient & midpoint

A

(2,3)

and

B

(8,11)

. Find (a) the gradient of

AB

(b) the length of

AB

AB

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-4024-CG1-0012 marksL1 FoundationCalculatorGradient & midpoint

Find the gradient of the line through

(1,2)

and

(4,11)

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG1-0022 marksL1 FoundationCalculatorGradient & midpoint

Find the midpoint of the line segment joining

(-2,5)

and

(6,-1)

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG-0072 marksL1 FoundationNon-calculatorGradient & midpoint

Find the distance between

(-3,2)

and

(5,-4)

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG-0012 marksL1 FoundationNon-calculatorStraight lines

Find the equation of the line through

(1,-2)

with gradient

3

. Give your answer in the form

y=mx+c

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG1-0042 marksL1 FoundationCalculatorStraight lines

The line through

(2,7)

with gradient

-1

has equation

y=-x+c

. Find

c

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG-0023 marksL2 StandardNon-calculatorStraight lines

Find the equation of the line through

(-1,4)

and

(3,-4)

, in the form

y=mx+c

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG-0032 marksL2 StandardNon-calculatorStraight lines

For the line

2x+3y=12

, find (a) the gradient (b) the

y

-intercept.

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG1-0031 markL1 FoundationCalculatorStraight lines

The line

y=3x-4

crosses the

x

-axis at

(k,0)

. Find

k

Salam Academy Original Practice · Proposed mark scheme

SA-4024-CG-0143 marksL2 StandardNon-calculatorStraight lines

The line

y=mx+c

passes through

(2,5)

and

(6,7)

. Find

m

and

c

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-4024-CG-0153 marksL3 ExamNon-calculatorGradient & midpoint

The points

(1,k)

(3,7)

and

(5,11)

lie on a straight line. Find

k

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Gradient $=\dfrac{\text{rise}}{\text{run}}=\dfrac{y_2-y_1}{x_2-x_1}$ .
Midpoint: average the coordinates. Length: Pythagoras.
$y=mx+c$ : gradient $m$ , intercept $c$ ; find $c$ by substituting a point.
Rearrange to $y=\ldots$ to read off $m$ and $c$ .

Section 12 of 14

Quick quiz

Q1 · SA-4024-CG-Q021 markL1 FoundationCalculatorStraight lines

The line

y=5-2x

crosses the

y

-axis at

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-4024-CG1-0012 marksL1 FoundationCalculatorGradient & midpoint

Find the gradient of the line through

(1,2)

and

(4,11)

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: GCSE Mathematics 8300
Where: Algebra: graphs
Expected: Equations of straight lines from a point and gradient or two points; midpoint and length.
Calculator: Paper 1 non-calculator; Papers 2 and 3 calculator.
Typical style: Problem-solving questions often combine topics.
Open in: 8300 · Papers 1–3 8300 · Papers 1–3

Section 14 of 14

Past paper questions

5 questions (34 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

4024/12 · Oct/Nov 2025Question 85 marks
Coordinates, a gradient and the missing vertex of a parallelogram.
Question paper, page 6 Mark scheme
4024/22 · Oct/Nov 2025Question 189 marks
Gradients, parallel lines and the intercept of a perpendicular bisector.
Question paper, page 12 Mark scheme
4024/12 · May/June 2025Question 175 marks
Rearrange a linear equation and find a perpendicular line.
Question paper, page 15 Mark scheme
4024/12 · Oct/Nov 2024Question 245 marks
Midpoint, gradient and a perpendicular line.
Question paper, page 16 Mark scheme
4024/22 · May/June 2024Question 1110 marks
Distance between points, dividing a line and a perpendicular bisector.
Question paper, page 18 Mark scheme

All 4024 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: AQA 8300, First teaching 2015 (official specification).