J560 · 7.01–7.04GraphsFoundation & HigherGrades 4–7

Straight Line Graphs

Q: What is the gradient of a vertical line?

It is undefined, because the change in x is zero and you cannot divide by zero.

Q: How do I plot y = mx + c quickly?

Plot the intercept (0, c), then move 1 across and m up (or down) to get a second point. Use a third point as a check.

Every straight line on a coordinate grid can be described by an equation of the form $y = mx + c$ . The number $m$ says how steep the line is, and $c$ says where it crosses the $y$ -axis. Once you can move fluently between a line, its equation and the points on it, a whole family of questions — plotting, finding equations, parallel and perpendicular lines — becomes routine.

55 min Graphs

The big picture

The gradient measures how much $y$ changes for each step of 1 in $x$ . That makes it a rate: on a distance–time graph it is a speed, on a cost graph it is a price per item. Reading $m$ and $c$ in context is a common OCR question in its own right.

Parallel lines share a gradient; perpendicular lines have gradients whose product is $-1$ . That single fact about perpendicular gradients is a Higher-tier favourite and connects directly to tangents to circles later in this chapter.

GCSE Maths - What on Earth is y = mx + c (2026/27 exams)Curated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

Before you start

Try a short calculation from the preceding topic without the notes.
Check the tier and grade band shown in this lesson.

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Plot straight line graphs from their equations

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: What is the gradient of $y = 5 - 2x$ ?

Show answer and reasoning

$-2$ . The coefficient of $x$ is $-2$ .

After watching 2: Find the gradient between $(1, 3)$ and $(5, 11)$ .

Show answer and reasoning

$2$ . $\tfrac{8}{4}$ .

What you'll learn

Plot straight line graphs from their equations
Find the gradient and $y$ -intercept from $y = mx + c$ and from a graph
Find the equation of a line through two points, or through a point with a given gradient
Identify and find parallel and perpendicular lines (perpendicular: Higher)

Gradient and intercept

In $y = mx + c$ , $m$ is the $\textbf{gradient}$ and $c$ is the $\textbf{y-intercept}$ , where the line crosses the $y$ -axis at $(0, c)$ .

A positive gradient slopes upwards from left to right; a negative gradient slopes downwards. A gradient of 0 gives a horizontal line $y = c$ ; vertical lines have the form $x = k$ .

If the equation is not in the form $y = mx + c$ , rearrange it first: $2y - 6x = 8$ becomes $y = 3x + 4$ .

y = mx + c

Gradient m, y-intercept c.

Your turnSame method, new numbers — try it before moving on.

What is the gradient of the line $3y = 6 - 2x$ ?

Your answer

Tip — The gradient is the coefficient of $x$ only after $y$ is the subject. In $2y = 6x + 8$ the gradient is 3, not 6.

Gradient between two points

The gradient between $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is the change in $y$ divided by the change in $x$ .

Subtract in the same order on the top and bottom. A drawn right-angled triangle on a graph gives the same value — rise over run.

m = \dfrac{y_{2} - y_{1}}{x_{2} - x_{1}}

Gradient between two points.

Example 1Find the equation of the line through

(-2, 5)

and

(4, -7)

m = \dfrac{-7 - 5}{4 - (-2)} = \dfrac{-12}{6} = -2

2Substitute

(-2, 5)

into

y = -2x + c

5 = 4 + c

, so

c = 1

Answer

y = -2x + 1

Your turnSame method, new numbers — try it before moving on.

Find the equation of the line through $(1, 3)$ and $(4, 12)$ .

Your answer

Type it like y = 2x + 1.

Parallel and perpendicular lines

$\textbf{Parallel}$ lines have equal gradients: $y = 3x - 2$ and $y = 3x + 7$ are parallel.

$\textbf{Perpendicular}$ lines have gradients that multiply to $-1$ . If one gradient is $m$ , the other is $-\dfrac{1}{m}$ — the negative reciprocal.

To find a line perpendicular to a given one through a point, take the negative reciprocal gradient and substitute the point to find $c$ .

m_{1} \times m_{2} = -1

Perpendicular gradients.

Example 1Find the equation of the line perpendicular to

y = 3x - 4

that passes through

(6, 1)

1Perpendicular gradient:

-\tfrac13

2Substitute

(6, 1)

1 = -2 + c

, so

c = 3

Answer

y = -\tfrac13 x + 3

Your turnSame method, new numbers — try it before moving on.

Find the equation of the line perpendicular to $y = -2x + 5$ that passes through $(4, 1)$ .

Your answer

Type it like y = 1/2x - 1.

Why this works

Rotating a line through $90^{\circ}$ swaps the rise and the run and changes one sign. A gradient of $\tfrac{3}{1}$ becomes $\tfrac{-1}{3}$ — which is exactly the negative reciprocal.

Lines in context

When a straight line models a real situation, $c$ is the starting value and $m$ is the rate of change.

A taxi fare of $F = 2.40d + 3.50$ pounds has a fixed charge of £3.50 and costs £2.40 per mile.

Example 1A plumber charges £45 for a 1-hour job and £105 for a 4-hour job. Assuming a linear charge, find the call-out fee and the hourly rate.

1Gradient:

\dfrac{105 - 45}{4 - 1} = 20

2Using

(1, 45)

45 = 20 + c

, so

c = 25

AnswerCall-out fee £25, plus £20 per hour.

Tip — In context questions, always interpret $m$ and $c$ with units: "£20 per hour", not just "20".

Think like an examiner

Show the gradient calculation with the coordinates substituted.
Show the substitution used to find $c$ .
Give equations in the form requested, such as $ax + by + c = 0$ with integers.

Straight lines

y = mx + c

Equation of a line.

m = \tfrac{y_{2} - y_{1}}{x_{2} - x_{1}}

Gradient.

m_{1}m_{2} = -1

Perpendicular lines.

Watch out for these

Computing

\dfrac{x_{2} - x_{1}}{y_{2} - y_{1}}

Change in

y

over change in

x

Taking the gradient of

3y = 6x + 2

as 6.

Make

y

the subject: gradient 2.

Using

-m

as the perpendicular gradient.

Use the negative reciprocal,

-\tfrac{1}{m}

Writing

y = 4

as a vertical line.

y = 4

is horizontal; vertical lines are

x = k

Stretch yourself

Points $A(1, 2)$ and $B(7, 6)$ are the ends of a line segment. Find the equation of the perpendicular bisector of $AB$ — the line through its midpoint at right angles to it.

Hint — Find the midpoint and the gradient of $AB$ , then use the negative reciprocal gradient through the midpoint.

Questions students ask

What is the gradient of a vertical line?

It is undefined, because the change in

x

is zero and you cannot divide by zero.

How do I plot

y = mx + c

quickly?

Plot the intercept

(0, c)

, then move 1 across and

m

up (or down) to get a second point. Use a third point as a check.

Do perpendicular lines always have one positive and one negative gradient?

Yes, unless one is horizontal and the other vertical.

Key takeaways

$y = mx + c$ : gradient $m$ , $y$ -intercept $(0, c)$ .
Gradient $= \tfrac{\text{change in } y}{\text{change in } x}$ .
Find $c$ by substituting a known point.
Parallel lines: equal gradients. Perpendicular lines: $m_{1}m_{2} = -1$ .
In context, $m$ is a rate and $c$ a starting value.

Exam-style questions

Practise it the way it is examined

3 original questions · 7 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 42 marks

Find the gradient and $y$ -intercept of the line $2y - 4x = 10$ .[2]

Question 2Grade 62 marks

Line $L$ passes through $(2, 7)$ and is parallel to $y = 3x - 1$ . Find the equation of $L$ .[2]

Question 3Grade 73 marks

$A$ is $(-1, 4)$ and $B$ is $(5, 2)$ . Find the equation of the perpendicular bisector of $AB$ .[3]

Unlimited practice

Skill drill

A new question every time, marked instantly, with a full worked solution. Aim for a streak of five, then move up a level.

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OCR J560 official specification (opens in a new tab)
Check the assessed scope and practical or mathematical requirements.
Read or explore this topic (opens in a new tab)
An external explanation or interactive activity to use alongside the lesson. It may include extension material.

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Which line is parallel to $y = 4x + 1$ ?
What is the gradient of $3x + 2y = 12$ ?
What gradient is perpendicular to $y = -\tfrac25 x + 3$ ?
Find the line through $(2, 7)$ perpendicular to $y = 2x$ .

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

Browse practice topics

Related lessons

All 5 lessons

Straight Line Graphs

Every straight line on a coordinate grid can be described by an equation of the form

y = mx + c

. The number

m

says how steep the line is, and

c

says where it crosses the

y

-axis. Once you can move fluently between a line, its equation and the points on it, a whole family of questions — plotting, finding equations, parallel and perpendicular lines — becomes routine.

55 min Graphs

Watch, pause, try

Video by Cognito. The video’s own board label is retained; follow the specification reference and scope notes on this lesson for assessed content.

Viewing focus: Plot straight line graphs from their equations

After watching, attempt these questions before opening the answers. You can also use the written lesson below.

After watching 1: What is the gradient of $y = 5 - 2x$ ?

Show answer and reasoning

$-2$ . The coefficient of $x$ is $-2$ .

After watching 2: Find the gradient between $(1, 3)$ and $(5, 11)$ .

Show answer and reasoning

$2$ . $\tfrac{8}{4}$ .

Practise it the way it is examined

Question 1Grade 42 marks

Find the gradient and $y$ -intercept of the line $2y - 4x = 10$ .[2]

Question 2Grade 62 marks

Line $L$ passes through $(2, 7)$ and is parallel to $y = 3x - 1$ . Find the equation of $L$ .[2]

Question 3Grade 73 marks

$A$ is $(-1, 4)$ and $B$ is $(5, 2)$ . Find the equation of the perpendicular bisector of $AB$ .[3]

Return to this tomorrow

Close the notes and answer from memory. Revisit any explanation you cannot yet reconstruct.

Which line is parallel to

y = 4x + 1

What is the gradient of

3x + 2y = 12

What gradient is perpendicular to

y = -\tfrac25 x + 3

Find the line through

(2, 7)

perpendicular to

y = 2x