What is the general form of the equation of a line in coordinate geometry?

In coordinate geometry, the general form of the equation of a line is expressed as Ax + By + C = 0, where A, B, and C are constants. This form is useful for analyzing the properties of the line, such as its slope and intercepts, which are important concepts in the Cambridge IGCSE Mathematics curriculum.

How do you find the slope of a line given two points?

To find the slope of a line given two points, (x1, y1) and (x2, y2), you use the formula: slope (m) = (y2 - y1) / (x2 - x1). This formula calculates the rate of change between the two points, which is a key concept in coordinate geometry and is often tested in the Cambridge IGCSE Mathematics exams.

What is the slope-intercept form of a line's equation and how is it used?

The slope-intercept form of a line's equation is y = mx + c, where m represents the slope and c represents the y-intercept. This form is particularly useful for quickly graphing a line and understanding its steepness and position on the y-axis, which are essential skills in the Cambridge IGCSE Mathematics syllabus.

How can you determine if two lines are parallel or perpendicular?

Two lines are parallel if their slopes are equal, meaning they have the same steepness and will never intersect. They are perpendicular if the product of their slopes is -1, indicating they intersect at a right angle. Understanding these relationships is crucial for solving problems in coordinate geometry, as covered in the Cambridge IGCSE Mathematics curriculum.

How do you find the equation of a line given a point and a slope?

To find the equation of a line given a point (x1, y1) and a slope m, you can use the point-slope form: y - y1 = m(x - x1). This form is particularly useful for constructing the equation of a line when specific points and slopes are provided, a common task in Cambridge IGCSE Mathematics problems.

Why is "Forgetting to divide all terms when isolating $y$" a common mistake in Equation of a Line?

Why it happens: 2y = -4x + 10 gets simplified to y = -4x + 5 instead of y = -2x + 5. How to avoid it: Divide EVERY term by the coefficient of y.

Why is "Using the reciprocal without changing sign" a common mistake in Equation of a Line?

Why it happens: Half-remembering the rule. How to avoid it: Perpendicular = NEGATIVE reciprocal: flip AND change sign. Source: 0580/42 — recurring

Why is "Saying parallel lines have the same equation" a common mistake in Equation of a Line?

Why it happens: Confusing "same gradient" with "same equation". How to avoid it: Parallel = same m, generally different c. The lines are distinct unless they're identical.

Why is "Computing the gradient of a vertical line as $\infty$ instead of saying "undefined"" a common mistake in Equation of a Line?

Why it happens: Vertical lines have x = 0, division by zero. How to avoid it: Vertical line x = k has UNDEFINED gradient. Horizontal line y = k has gradient 0.

Cambridge IGCSE Mathematics (0580)

Equation of a Line

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Short Notes - Equation of a Line

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Detailed Study Notes

Detailed notes on Coordinate Geometry for Cambridge IGCSE Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Equation of a Straight Line — Cambridge IGCSE 0580 Maths Extended (2026)

Two reliable forms — $y = mx + c$ and $y - y_{1} = m(x - x_{1})$ — let you write the equation of any line from a point and a gradient, or from two points. Cambridge tests this in nearly every Paper 4 coordinate-geometry question.

At a glance

Slope-intercept: $y = mx + c$ .
Point-gradient: $y - y_{1} = m(x - x_{1})$ .
From two points: find $m$ , then use point-gradient with EITHER point.
Rearrange any form to $y = mx + c$ to read off gradient and $y$ -intercept.
Parallel: same $m$ . Perpendicular: $m_{2} = -\dfrac{1}{m_{1}}$ .
Horizontal line: $y = c$ . Vertical line: $x = a$ .

What you’ll learn

Mapped to the Cambridge IGCSE 0580 syllabus (2025-2027).

E3.2 — Interpret and obtain the equation of a straight-line graph in the form $y = mx + c$ .
Find the equation of a line through given points or with given gradient and a point.

Slope-intercept form

$y = mx + c$ . Read off gradient and $y$ -intercept directly.

$y = mx + c$ .

$m$ = gradient (slope).
$c$ = $y$ -intercept (where the line crosses the $y$ -axis).

To plot: mark the $y$ -intercept at $(0, c)$ , then use the gradient to step rise/run for a second point.

c is where the line cuts the y-axis; m is the rise over the run of any step along it.

Worked. Plot $y = -2x + 5$ .

$y$ -intercept at $(0, 5)$ .
Gradient $-2$ means: every $1$ right, go $2$ down. So $(1, 3)$ , $(2, 1)$ , etc.

To find $c$ given a point and gradient. Substitute.

Worked. Line has gradient $3$ and passes through $(2, 7)$ . Find $c$ .

$7 = 3(2) + c \Rightarrow c = 1$ .
Equation: $y = 3x + 1$ .

$m$ = gradient.
$c$ = $y$ -intercept.
Substitute one point to find $c$ if gradient is known.

Point-gradient form

$y - y_{1} = m(x - x_{1})$ . The fastest way when you have one point and a gradient.

Formula. $y - y_{1} = m(x - x_{1})$ .

This works when you know:

A specific point $(x_{1}, y_{1})$ ON the line.
The gradient $m$ .

Worked. Line with gradient $4$ through $(2, -3)$ .

$y - (-3) = 4(x - 2)$ .
$y + 3 = 4x - 8$ .
$y = 4x - 11$ (rearranged to slope-intercept).

Worked. Line through $(-1, 5)$ with gradient $-\dfrac{1}{2}$ .

$y - 5 = -\dfrac{1}{2}(x - (-1))$ .
$y - 5 = -\dfrac{1}{2}(x + 1)$ .
$y = -\dfrac{1}{2}x - \dfrac{1}{2} + 5 = -\dfrac{1}{2}x + \dfrac{9}{2}$ .

When the gradient is a fraction, you may want to multiply through to clear fractions.

$y - y_{1} = m(x - x_{1})$ .
Best when you have a point + a gradient.
Rearrange to $y = mx + c$ for the final form.

From two points

Compute the gradient, then plug into point-gradient form using either point.

Method.

Compute the gradient $m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .
Use point-gradient form with EITHER given point.
Rearrange.

Worked. Line through $A(1, 4)$ and $B(5, 12)$ .

$m = \dfrac{12 - 4}{5 - 1} = \dfrac{8}{4} = 2$ .
Using $A$ : $y - 4 = 2(x - 1) \Rightarrow y = 2x + 2$ .
Check with $B$ : $2(5) + 2 = 12$ ✓.

Find $m$ from the two points.
Use point-gradient with EITHER point — same final equation.
Always check the OTHER point satisfies your equation.

Parallel and perpendicular lines

Parallel: same gradient. Perpendicular: negative reciprocal.

Parallel. Two lines are parallel if and only if they have the SAME gradient.

Worked. Line parallel to $y = 3x + 1$ through $(2, 4)$ .

Gradient: $3$ .
$y - 4 = 3(x - 2) \Rightarrow y = 3x - 2$ .

Perpendicular. Two lines are perpendicular when their gradients multiply to $-1$ . Each is the negative reciprocal of the other.

Parallel lines share a gradient; perpendicular lines have gradients that multiply to −1.

Worked. Line perpendicular to $y = 2x + 5$ through $(0, 3)$ .

Original gradient: $2$ .
Perpendicular gradient: $-\dfrac{1}{2}$ .
$y - 3 = -\dfrac{1}{2}(x - 0) \Rightarrow y = -\dfrac{1}{2}x + 3$ .

Special case. If a line is vertical ( $x = a$ ), the perpendicular is horizontal ( $y = b$ ). Their gradients are undefined and zero respectively — the " $m_{1} \cdot m_{2} = -1$ " rule doesn't apply directly.

Parallel: same gradient.
Perpendicular: $m_{2} = -\dfrac{1}{m_{1}}$ .
Vertical and horizontal are perpendicular (special case).
Use point-gradient form with the given point and the new gradient.

Horizontal and vertical lines

$y = c$ is horizontal. $x = a$ is vertical. Don't try to put them in $y = mx + c$ form.

Horizontal line $y = c$ . Constant $y$ . Gradient $0$ . Cannot use $y = mx + c$ in the usual way (it works, with $m = 0$ , but the equation is just $y = c$ ).

Vertical line $x = a$ . Constant $x$ . Gradient undefined. CANNOT be written as $y = mx + c$ — the slope-intercept form simply doesn't apply.

Recognising them.

"Find the line through $(3, 4)$ parallel to the $y$ -axis" → vertical → $x = 3$ .
"Find the line through $(3, 4)$ parallel to the $x$ -axis" → horizontal → $y = 4$ .

Horizontal: $y = c$ (gradient $0$ ).
Vertical: $x = a$ (gradient undefined).
Vertical can't be written as $y = mx + c$ .

Quick recap

$y = mx + c$ : gradient and $y$ -intercept directly.
$y - y_{1} = m(x - x_{1})$ : when you have a point and a gradient.
Two points: find $m$ , then use point-gradient.
Parallel: same $m$ . Perpendicular: $-1/m$ .
Horizontal: $y = c$ . Vertical: $x = a$ .

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Slope-intercept form — $y = mx + c$ .
Point-gradient form — $y - y_{1} = m(x - x_{1})$ .
Parallel lines — same gradient.
Perpendicular lines — gradients multiply to $-1$ .

How it’s examined

Equation-of-a-line questions appear on every Paper 4 — typically 4-6 marks combining gradient, point, and parallel/perpendicular logic. Paper 2 has 2-3 mark single-line items. Examiner reports flag forgetting to substitute the point's coordinates correctly and computing $-1/m$ wrongly.

Sources: Cambridge IGCSE Mathematics 0580 syllabus 2025-2027 (E3.2); 0580/42 Oct/Nov 2024 — Q8 (perpendicular line equation); 0580 Examiner Reports 2022-2024. Last reviewed 2026-05-05.

Take this whole topic with you

Step-by-step worked examples — Equation of a Line

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

Question type:

Question patterns to master — Equation of a Line

Almost every equation of a line exam question is one of these shapes. Learn to spot each one and you will always know how to start.

Direct calculation

Recognise it by

A single instruction — find the equation of the line from a point and gradient, two points, or a parallel/perpendicular condition; or state the gradient and intercept.

How to approach it

Get the gradient (given, computed from two points, or copied/negated-reciprocated for parallel/perpendicular), then substitute into $y - y_{1} = m(x - x_{1})$ and rearrange into $y = mx + c$ .

Common trap

Using the reciprocal without negating it for a perpendicular line, or failing to divide every term when isolating $y$ . Examiner reports flag both as recurring rearrangement slips.

Multi-step problem

Recognise it by

Several stages chained — find a midpoint then a line, find the intersection of two lines, or build a perpendicular bisector.

How to approach it

Work the stages in order: midpoint first, then gradient, then the perpendicular gradient, then the line. For an intersection, substitute one equation into the other and solve.

Common trap

Passing the perpendicular bisector through $A$ or $B$ instead of the midpoint $M$ . Examiner reports flag this — the bisector must go through $M$ .

Show that / prove

Recognise it by

The words show that — typically that a point equidistant from two others lies on a given line (the perpendicular bisector).

How to approach it

Set the squared distances equal to avoid surds, expand both sides, cancel the $x^{2}$ and $y^{2}$ terms, and simplify to the stated equation.

Common trap

Working with un-squared distances and getting tangled in square roots. Examiner reports note the distance-squared form is what earns the method marks.

1Equation from gradient and a point
CoreDirect calculation• point-gradient form
Question
Find the equation of the line with gradient $3$ passing through $(2, 5)$ .
Step-by-step solution
1. Step 1
  Use $y - y_{1} = m(x - x_{1})$ .
  $y - 5 = 3(x - 2)$
2. Step 2
  Expand and simplify.
  $y = 3x - 1$
Answer
$y = 3x - 1$
2Equation from two points
ExtendedDirect calculation• Adapted from 0580/42 May/Jun 2024 Q11• two points
Question
Find the equation of the line passing through $A(-1, 4)$ and $B(3, -8)$ .
Step-by-step solution
1. Step 1
  Find the gradient.
  $m = \dfrac{-8 - 4}{3 - (-1)} = \dfrac{-12}{4} = -3$
2. Step 2
  Use point-gradient form with $A$ .
  $y - 4 = -3(x + 1)$
3. Step 3
  Expand.
  $y = -3x + 1$
Answer
$y = -3x + 1$
3Find a line parallel to a given one
ExtendedDirect calculation• parallel
Question
Find the equation of the line through $(4, 7)$ that is parallel to $y = 2x - 5$ .
Step-by-step solution
1. Step 1
  Parallel → same gradient $m = 2$ .
2. Step 2
  Use point-gradient form.
  $y - 7 = 2(x - 4) \implies y = 2x - 1$
Answer
$y = 2x - 1$
4Find a line perpendicular to a given one
ExtendedDirect calculation• Adapted from 0580/42 Oct/Nov 2024 Q9• perpendicular
Question
Find the equation of the line through $(2, 6)$ that is perpendicular to $y = -\dfrac{1}{3}x + 4$ .
Step-by-step solution
1. Step 1
  Original gradient is $-\frac{1}{3}$ . Perpendicular = negative reciprocal.
  $m_{\perp} = 3$
2. Step 2
  Apply point-gradient form.
  $y - 6 = 3(x - 2) \implies y = 3x$
Answer
$y = 3x$
5Rearrange a general form into $y = mx + c$
CoreDirect calculation• rearrange
Question
State the gradient and y-intercept of $4x + 2y = 10$ .
Step-by-step solution
1. Step 1
  Solve for $y$ .
  $2y = -4x + 10 \implies y = -2x + 5$
2. Step 2
  Read off.
  $m = -2,\ c = 5$
Answer
Gradient $-2$ ; y-intercept $5$
6Find the intersection of two lines
ExtendedMulti-step problem• Adapted from 0580/22 Oct/Nov 2023 Q14• intersection, simultaneous
Question
Find the point of intersection of $y = 3x - 4$ and $2x + y = 11$ .
Step-by-step solution
1. Step 1
  Substitute the first equation into the second.
  $2x + (3x - 4) = 11$
2. Step 2
  Simplify.
  $5x - 4 = 11 \implies x = 3$
3. Step 3
  Substitute back.
  $y = 3(3) - 4 = 5$
Answer
$(3, 5)$
7Equation of a horizontal or vertical line
ExtendedDirect calculation• horizontal, vertical
Question
Write the equation of (a) the horizontal line through $(2, -5)$ and (b) the vertical line through $(7, 3)$ .
Step-by-step solution
1. Step 1
  (a) Horizontal lines have the form $y = c$ , where $c$ is the y-coordinate.
  $y = -5$
2. Step 2
  (b) Vertical lines have the form $x = k$ , where $k$ is the x-coordinate.
  $x = 7$
Answer
(a) $y = -5$ (b) $x = 7$
Examiner tip
The examiner report flags candidates often swap $x$ and $y$ for these two cases. Memorise: HORIZONTAL → $y = \ldots$ , VERTICAL → $x = \ldots$ .
8Line through the midpoint of two points
ExtendedMulti-step problem• midpoint, two points
Question
Find the equation of the line passing through the midpoint of $A(2, 4)$ and $B(8, -2)$ with gradient $-3$ .
Step-by-step solution
1. Step 1
  Midpoint of $AB$ .
  $M = \left(\dfrac{2 + 8}{2}, \dfrac{4 + (-2)}{2}\right) = (5, 1)$
2. Step 2
  Apply point-gradient form.
  $y - 1 = -3(x - 5)$
3. Step 3
  Expand.
  $y = -3x + 16$
Answer
$y = -3x + 16$
9Find the perpendicular bisector
ChallengeMulti-step problem• Adapted from 0580/42 May/Jun 2024 Q14• perpendicular bisector, midpoint
Question
Find the equation of the perpendicular bisector of the segment from $A(1, 2)$ to $B(7, 6)$ .
Step-by-step solution
1. Step 1
  Find the midpoint.
  $M = (4, 4)$
2. Step 2
  Gradient of $AB$ .
  $m_{AB} = \dfrac{6 - 2}{7 - 1} = \dfrac{2}{3}$
3. Step 3
  Perpendicular gradient.
  $m_{\perp} = -\dfrac{3}{2}$
4. Step 4
  Apply point-gradient form through $M$ .
  $y - 4 = -\dfrac{3}{2}(x - 4) \implies y = -\dfrac{3}{2}x + 10$
Answer
$y = -\dfrac{3}{2}x + 10$
Examiner tip
The mark scheme awards method marks for the midpoint, the original gradient and the perpendicular gradient separately. The examiner report flags candidates often forget the perpendicular bisector passes through $M$ , not $A$ or $B$ .
10Find a line equidistant from two points
ChallengeShow that / prove• perpendicular bisector, equidistant
Question
A point $P(x, y)$ is equidistant from $A(0, 0)$ and $B(6, 4)$ . Show that $P$ lies on the line $3x + 2y = 13$ .
Step-by-step solution
1. Step 1
  Equidistant means $P$ lies on the perpendicular bisector of $AB$ . Use squared distances to avoid surds.
  $x^{2} + y^{2} = (x - 6)^{2} + (y - 4)^{2}$
2. Step 2
  Expand the right side.
  $x^{2} + y^{2} = x^{2} - 12x + 36 + y^{2} - 8y + 16$
3. Step 3
  Cancel and simplify.
  $0 = -12x - 8y + 52 \implies 12x + 8y = 52$
4. Step 4
  Divide by $4$ .
  $3x + 2y = 13$
Answer
$P$ lies on $3x + 2y = 13$ (the perpendicular bisector of $AB$ ).
Examiner tip
The examiner report flags candidates often forget to square both distances to avoid square roots. The mark scheme awards method marks for the equidistance equation in distance-squared form.

Key Formulae — Equation of a Line

The formulae you need to memorise for equation of a line on the Cambridge IGCSE 0580 paper, with every variable defined in plain English and a note on when to use it.

Slope-intercept form
$y = mx + c$
$m$
gradient
$c$
y-intercept
When to use
Standard form for a straight line — easiest to graph and read off.
Point-gradient form
$y - y_{1} = m(x - x_{1})$
$m$
gradient
$(x_{1}, y_{1})$
any known point on the line
When to use
Quickest way to write a line equation when you know a point and the slope.
General form
$ax + by + c = 0$
When to use
Some answers are required in this form — rearrange to read off $m$ and $c$ .

Key Definitions and Keywords — Equation of a Line

Definitions to memorise and the exact keywords mark schemes credit for equation of a line answers — sharpened from recent examiner reports for the 2026 0580 sitting.

Equation of a line
Examiner keyword
An algebraic relationship between $x$ and $y$ that every point on the line satisfies.
y-intercept
Examiner keyword
The y-coordinate of the point where the line crosses the y-axis ( $x = 0$ ).
x-intercept
The x-coordinate where the line crosses the x-axis ( $y = 0$ ).
Collinear points
Three or more points lying on the same straight line.

Common Mistakes and Misconceptions — Equation of a Line

The traps other students keep falling into on equation of a line questions — taken from recent Cambridge IGCSE 0580 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to divide all terms when isolating $y$
Why it happens
$2y = -4x + 10$ gets simplified to $y = -4x + 5$ instead of $y = -2x + 5$ .
How to avoid it
Divide EVERY term by the coefficient of $y$ .
✕Using the reciprocal without changing sign
0580/42 — recurring
Why it happens
Half-remembering the rule.
How to avoid it
Perpendicular = NEGATIVE reciprocal: flip AND change sign.
✕Saying parallel lines have the same equation
Why it happens
Confusing "same gradient" with "same equation".
How to avoid it
Parallel = same $m$ , generally different $c$ . The lines are distinct unless they're identical.
✕Computing the gradient of a vertical line as $\infty$ instead of saying "undefined"
Why it happens
Vertical lines have $\Delta x = 0$ , division by zero.
How to avoid it
Vertical line $x = k$ has UNDEFINED gradient. Horizontal line $y = k$ has gradient $0$ .