What is a linear graph in the context of Edexcel IGCSE Mathematics?

In Edexcel IGCSE Mathematics, a linear graph represents a straight line on a coordinate plane. It is defined by a linear equation of the form y = mx + c, where m is the gradient (slope) and c is the y-intercept. Linear graphs are fundamental in understanding relationships between variables in sequences, functions, and graphs.

How do you find the gradient of a linear graph?

The gradient of a linear graph, often referred to as the slope, measures the steepness of the line. It is calculated by the formula m = (change in y) / (change in x), or rise over run. This can be determined by selecting two points on the line and dividing the difference in their y-coordinates by the difference in their x-coordinates.

What is the significance of the y-intercept in a linear equation?

The y-intercept of a linear equation is the point where the graph crosses the y-axis. In the equation y = mx + c, the y-intercept is represented by c. It indicates the value of y when x is zero, providing a starting point for graphing the line and understanding its position on the coordinate plane.

How can you determine if two linear graphs are parallel?

Two linear graphs are parallel if they have the same gradient but different y-intercepts. This means their equations will have the same value for m in y = mx + c, but different values for c. Parallel lines never intersect and maintain a constant distance apart.

What is the process for graphing a linear equation on a coordinate plane?

To graph a linear equation on a coordinate plane, start by identifying the y-intercept (c) and plot it on the y-axis. Then, use the gradient (m) to determine the direction and steepness of the line. From the y-intercept, move vertically and horizontally according to the gradient to plot a second point. Draw a straight line through these points to complete the graph.

Why is "Computing gradient as $(x_2 - x_1)/(y_2 - y_1)$" a common mistake in Linear Graphs?

Why it happens: Reversed. How to avoid it: Gradient = RISE OVER RUN = y / x. y on top.

Why is "Using just reciprocal (without negative) for perpendicular" a common mistake in Linear Graphs?

Why it happens: Halfway. How to avoid it: Perpendicular = NEGATIVE RECIPROCAL. m -1/m. Both flip AND negate. Source: 4MA1/2H May/Jun 2024 — examiner report Q15

Why is "Adding coordinates without dividing by 2" a common mistake in Linear Graphs?

Why it happens: Forgetting the average. How to avoid it: Midpoint = AVERAGE of coordinates. Add THEN divide by 2.

Sequences, Functions and GraphsPearson Edexcel IGCSE Mathematics 4MA1 Higher

Linear Graphs

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next.

Previous Next

Learn this Topic

Start with the slides for the quick version, then go deeper with the full study notes.

Short Study Notes in the form of Slides

Read the notes first. If the method in a worked example clicks, you're ready for the questions.

Short Study Notes — Linear Graphs

Start with these resources to cover the key concepts, then work through the practice questions.

Page 1 / 0

Detailed Notes

Full prose, callouts and a recap — built for A* mastery, not just a quick scan.

Take these study notes with you

Download a branded PDF — full prose, callouts, recap and memorise list for Linear Graphs, ready to print or save offline.

Linear Graphs Study Notes — Edexcel IGCSE 4MA1 Higher Tier (2026 onwards)

Plot, read and find equations of straight lines. Gradient, intercept, parallel, perpendicular. Foundation for coordinate geometry.

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4MA1 syllabus (2026 onwards).

3.3 — Find the gradient and y-intercept of a line.
3.3 — Find the equation of a line through two points or a point and a gradient.
3.3 — Identify parallel and perpendicular lines.
3.3 — Find the midpoint of two points.

Equation of a straight line

$y = mx + c$ — gradient $m$ , y-intercept $c$ .

Standard form: $y = mx + c$ .

$m$ = gradient (slope, steepness).
$c$ = y-intercept (where it crosses $y$ -axis).

c is where the line cuts the y-axis; m is rise divided by run.

Reading from equation:

Equation	Gradient	y-intercept
$y = 3x + 5$	$3$	$5$
$y = -2x + 7$	$-2$	$7$
$y = \dfrac{1}{2}x - 3$	$\dfrac{1}{2}$	$-3$
$y = -x$	$-1$	$0$
$y = 4$	$0$	$4$ (horizontal line)

Vertical lines: $x = a$ . No gradient (or 'undefined').

Worked qualitative. What does a negative gradient mean?

Line slopes DOWN from left to right.
$y$ DECREASES as $x$ increases.
E.g. $y = -3x + 5$ : at $x = 0, y = 5$ ; at $x = 1, y = 2$ .

Edexcel tip. Always rewrite the equation in $y = mx + c$ form before reading off $m$ and $c$ .

$y = mx + c$ .
$m$ = gradient.
$c$ = y-intercept.
Negative gradient: down-sloping.

Finding the equation of a line

From two points, or point + gradient. Use $y - y_1 = m(x - x_1)$ .

Method 1: Two points.

Find gradient: $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ .
Use one point in $y - y_1 = m(x - x_1)$ .
Expand and simplify.

Example. Points $(2, 7)$ and $(5, 16)$ .

$m = (16 - 7) / (5 - 2) = 3$ .
$y - 7 = 3(x - 2)$ .
$y = 3x - 6 + 7 = 3x + 1$ .

Method 2: Point + gradient.

Just plug into $y - y_1 = m(x - x_1)$ .

Example. Through $(3, 4)$ with gradient $-2$ .

$y - 4 = -2(x - 3)$ .
$y = -2x + 6 + 4 = -2x + 10$ .

Method 3: Two intercepts.

If a line crosses both axes at known points, e.g. $(a, 0)$ and $(0, b)$ , equation: $\dfrac{x}{a} + \dfrac{y}{b} = 1$ .

Worked qualitative. Why is the point-slope form so useful?

Works regardless of whether $c$ is a 'nice' number.
Avoids solving for $c$ separately.
Single substitution gets the equation.

Edexcel tip. Show every step. Mark scheme: M1 for gradient, M1 for substituting point, A1 for the equation.

Two points: find gradient, use point-slope.
Point + gradient: direct substitution.
Always state $m$ explicitly.

Parallel and perpendicular

Parallel: same gradient. Perpendicular: gradients × = $-1$ .

Parallel: SAME gradient. y-intercepts can differ.

If line A has gradient $3$ , any line parallel has gradient $3$ too.

Example. Find the line parallel to $y = 2x - 5$ through $(3, 4)$ .

Same gradient: $m = 2$ .
$y - 4 = 2(x - 3)$ → $y = 2x - 2$ .

Perpendicular: gradients multiply to $-1$ .

If $m_1 \cdot m_2 = -1$ , then $m_2 = -1/m_1$ — the NEGATIVE RECIPROCAL.

Gradient $m_1$	Perpendicular $m_2$
$2$	$-\dfrac{1}{2}$
$3$	$-\dfrac{1}{3}$
$-\dfrac{1}{2}$	$2$
$-4$	$\dfrac{1}{4}$

Example. Find line perpendicular to $y = \dfrac{1}{2}x + 3$ through $(4, -1)$ .

Negative reciprocal of $\dfrac{1}{2}$ : $-2$ .
$y + 1 = -2(x - 4) \to y = -2x + 7$ .

Verify. $\dfrac{1}{2} \times (-2) = -1$ ✓.

Parallel lines share a gradient; perpendicular gradients are negative reciprocals.

Worked qualitative. Why does perpendicular need NEGATIVE reciprocal?

Lines at $90°$ .
Gradient = tan(angle from horizontal).
$\tan(\theta) \times \tan(\theta + 90°) = -1$ (trig identity).
Geometrically: rotate by $90°$ flips and negates.

Edexcel tip. When you say 'gradient is $-2$ because it's perpendicular to $y = \dfrac{1}{2}x + 3$ ', state the reasoning. Mark schemes credit it.

Parallel: same $m$ .
Perpendicular: $m_1 \cdot m_2 = -1$ .
Perpendicular = negative reciprocal.
Verify: product = $-1$ .

Midpoint of two points

Average of coordinates.

Formula: midpoint of $(x_1, y_1)$ and $(x_2, y_2)$ is:

$M = \left(\dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2}\right)$

Example. $A = (3, 7)$ , $B = (9, 1)$ .

$M = \left(\dfrac{3 + 9}{2}, \dfrac{7 + 1}{2}\right) = (6, 4)$ .

Common application: perpendicular bisector.

The perpendicular bisector of $AB$ :

Goes through the midpoint of $A$ and $B$ .
Is perpendicular to $AB$ .

Find:

Midpoint $M$ .
Gradient of $AB$ .
Negative reciprocal as gradient of bisector.
Use point-slope at $M$ .

Worked qualitative. Why average the coordinates for midpoint?

Halfway between two values: arithmetic mean.
Same for $x$ AND $y$ separately.
Geometrically: equidistant from both endpoints.

Edexcel tip. Show the formula explicitly: ' $M = ((3+9)/2, (7+1)/2)$ '. Earns the M1.

Midpoint: average each coordinate.
Used for perpendicular bisectors.
Always show the formula.

How it’s examined

Linear graphs every Higher Tier paper (4-7 marks). Often combined with perpendicular/parallel. Examiner reports flag (1) gradient formula reversed, (2) just reciprocal (no negative) for perpendicular, (3) forgetting to divide for midpoint.

Sources: Pearson Edexcel International GCSE Mathematics A 4MA1 specification (2026 onwards); 4MA1 Examiner Reports — Jan and May/Jun 2022-2024 series; Past Papers — 4MA1/1H Jan 2024, 4MA1/2H May/Jun 2024. Last reviewed 2026-05-07.

Master this Topic

Worked examples, formulae, definitions and the mistakes examiners flag — everything you need to push from a pass to an A*.

Take this whole topic with you

Download a branded revision sheet — worked examples, formulae, definitions and common mistakes for Linear Graphs, ready to print or save as PDF.

Step-by-step worked examples — Linear Graphs

Step-by-step solutions to past-paper-style questions on linear graphs, written exactly the way a tutor would explain them at the board.

1Find gradient and y-intercept
Foundation• gradient, intercept
Question
Find the gradient and y-intercept of $y = 3x - 5$ .
Step-by-step solution
1. Step 1
  Compare with $y = mx + c$ : $m = 3$ (gradient), $c = -5$ (y-intercept).
Answer
Gradient = $3$ , y-intercept = $(0, -5)$ .
2Equation from two points
Higher• Adapted from 4MA1/1H Jan 2024 Q10• gradient, two points
Question
Find the equation of the line through $(2, 7)$ and $(5, 16)$ .
Step-by-step solution
1. Step 1
  Gradient: $m = \dfrac{16 - 7}{5 - 2} = \dfrac{9}{3} = 3$ .
2. Step 2
  Use $y - y_1 = m(x - x_1)$ with point $(2, 7)$ : $y - 7 = 3(x - 2)$ .
3. Step 3
  Expand: $y = 3x - 6 + 7 = 3x + 1$ .
Answer
$y = 3x + 1$
Examiner tip
ALWAYS show the gradient calculation. Mark schemes give M1 for gradient, M1 for substitution, A1 for the equation.
3Parallel line through a point
Higher• parallel
Question
Find the equation of the line parallel to $y = 2x - 5$ that passes through $(3, 4)$ .
Step-by-step solution
1. Step 1
  Parallel lines have the SAME gradient: $m = 2$ .
2. Step 2
  Use $y - 4 = 2(x - 3)$ :
  $y = 2x - 6 + 4 = 2x - 2$
Answer
$y = 2x - 2$
4Perpendicular line
Higher• Adapted from 4MA1/2H May/Jun 2024 Q15• perpendicular
Question
Find the equation of the line perpendicular to $y = \dfrac{1}{2}x + 3$ passing through $(4, -1)$ .
Step-by-step solution
1. Step 1
  Perpendicular gradient: NEGATIVE RECIPROCAL of $\dfrac{1}{2}$ → $-2$ .
2. Step 2
  Use $y - (-1) = -2(x - 4)$ :
  $y + 1 = -2x + 8 \implies y = -2x + 7$
Answer
$y = -2x + 7$
Examiner tip
Perpendicular: gradients multiply to $-1$ . So $\dfrac{1}{2} \times (-2) = -1$ ✓.
5Midpoint of two points
Foundation• midpoint
Question
Find the midpoint of $A(3, 7)$ and $B(9, 1)$ .
Step-by-step solution
1. Step 1
  Midpoint: average each coordinate.
  $M = \left(\frac{3 + 9}{2}, \frac{7 + 1}{2}\right) = (6, 4)$
Answer
$(6, 4)$

Key Formulae — Linear Graphs

The formulae you need to memorise for linear graphs on the Pearson Edexcel IGCSE 4MA1 paper, with every variable defined in plain English and a note on when to use it.

Gradient between two points
$m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$
$(x_1, y_1), (x_2, y_2)$
two points on the line
When to use
Whenever you need the gradient and have two points.
Example
Points $(2, 7)$ and $(5, 16)$ : $m = (16 - 7)/(5 - 2) = 3$ .
Line equation (point-slope form)
$y - y_{1} = m(x - x_{1})$
$m$
gradient
$(x_1, y_1)$
any point on the line
When to use
When you have a point and gradient.
Example
$m = 3$ , $(2, 7)$ : $y - 7 = 3(x - 2) \to y = 3x + 1$ .
Perpendicular gradients
$m_{1} \times m_{2} = -1$
$m_1, m_2$
gradients of perpendicular lines
When to use
When finding equations of perpendicular lines.
Example
Perpendicular to $m = 1/2$ : $-2$ . (Negative reciprocal.)
Midpoint formula
$M = \left(\dfrac{x_{1} + x_{2}}{2}, \dfrac{y_{1} + y_{2}}{2}\right)$
When to use
Finding midpoint of two points.
Example
$(3, 7)$ and $(9, 1)$ : midpoint $(6, 4)$ .

Key Definitions and Keywords — Linear Graphs

Definitions to memorise and the exact keywords mark schemes credit for linear graphs answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IGCSE 4MA1 sitting.

Gradient (slope)
Examiner keyword
The steepness of a line. Rise over run: $m = \dfrac{\Delta y}{\Delta x}$ .
Example
Line $y = 3x + 2$ has gradient $3$ .
y-intercept
Examiner keyword
Where the line crosses the $y$ -axis. In $y = mx + c$ , the y-intercept is $c$ .
Example
$y = 3x - 5$ : y-intercept $-5$ , point $(0, -5)$ .
Parallel lines
Lines with the SAME gradient.
Example
$y = 2x + 5$ parallel to $y = 2x - 3$ .
Perpendicular lines
Examiner keyword
Lines whose gradients multiply to $-1$ .
Example
$y = 2x$ perpendicular to $y = -\dfrac{1}{2}x$ .

Common Mistakes and Misconceptions — Linear Graphs

The traps other students keep falling into on linear graphs questions — taken from recent Pearson Edexcel IGCSE 4MA1 examiner reports and mark schemes — and how to avoid them.

✕Computing gradient as $(x_2 - x_1)/(y_2 - y_1)$
Why it happens
Reversed.
How to avoid it
Gradient = RISE OVER RUN = $\Delta y / \Delta x$ . $y$ on top.
✕Using just reciprocal (without negative) for perpendicular
4MA1/2H May/Jun 2024 — examiner report Q15
Why it happens
Halfway.
How to avoid it
Perpendicular = NEGATIVE RECIPROCAL. $m \to -1/m$ . Both flip AND negate.
✕Adding coordinates without dividing by 2
Why it happens
Forgetting the average.
How to avoid it
Midpoint = AVERAGE of coordinates. Add THEN divide by 2.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

Get a report showing which sub-topics you've nailed and which ones still need work.

Video lesson

Short walkthrough of the concepts students most often get stuck on.

Linear Graphs — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Linear Graphs

Short Study Notes in the form of Slides

Short Study Notes — Linear Graphs

Detailed Notes

What you’ll learn

How it’s examined

1Find gradient and y-intercept

2Equation from two points

3Parallel line through a point

4Perpendicular line

5Midpoint of two points

Gradient between two points

Line equation (point-slope form)

Perpendicular gradients

Midpoint formula

Gradient (slope)

y-intercept

Parallel lines

Perpendicular lines

✕Computing gradient as (x2−x1)/(y2−y1)(x_2 - x_1)/(y_2 - y_1)(x2​−x1​)/(y2​−y1​)

✕Using just reciprocal (without negative) for perpendicular

✕Adding coordinates without dividing by 2

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Linear Graphs — frequently asked questions

✕Computing gradient as $(x_2 - x_1)/(y_2 - y_1)$