A8, A9, A10AlgebraFoundation & HigherGrades 3–7

Straight Line Graphs

Q: What does a gradient of 0 mean?

The line is horizontal — y does not change as x increases. Its equation is y = c.

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. Vertical lines have equations x = a.

Every straight line on a coordinate grid has an equation that can be written as $y = mx + c$ . The number $m$ is the gradient — how steep the line is — and $c$ is where it crosses the $y$ -axis. With those two facts you can sketch any line, find its equation from two points, and spot parallel and perpendicular lines instantly.

50 min Video by Maths Genie AQA GCSE Maths

The big picture

An equation such as $y = 2x + 1$ is a rule connecting the coordinates of every point on the line. Pick any $x$ , apply the rule, and the point $(x, y)$ is on the line. That is why a table of values works.

Gradient is a rate: how much $y$ changes for each 1 that $x$ increases. This idea returns in real-life graphs, where a gradient might be a speed in m/s or a cost per minute.

Drawing Straight Line GraphsCurated by Kepler, not produced by us. Watch it, then read the notes below.

Open on YouTube

What you'll learn

Plot straight line graphs from a table of values
Find the gradient and $y$ -intercept of a line from its graph or equation
Find the equation of a line through two points
Identify parallel and perpendicular lines (perpendicular: Higher)

Plotting from a table

Choose some $x$ values, substitute each into the equation to find $y$ , plot the points and join them with a ruler. Three points are enough — if they do not line up, one is wrong.

Lines of the form $x = a$ are vertical and $y = b$ are horizontal.

Example 1Complete a table of values for

y = 3x - 2

from

x = -1

x = 2

x = -1

y = -5

x = 0

y = -2

x = 1

y = 1

x = 2

y = 4

Answer

Points

(-1, -5)

(0, -2)

(1, 1)

(2, 4)

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

For $y = 5 - 2x$ , what is $y$ when $x = -3$ ?

Your answer

Show worked solution

Solution

1 $y = 5 - 2(-3) = 5 + 6 = 11$ .

Answer: $y = 11$

Gradient and intercept

Gradient $= \dfrac{\text{change in } y}{\text{change in } x}$ , often remembered as "rise over run". Lines going up from left to right have a positive gradient; lines going down have a negative gradient.

In $y = mx + c$ , $m$ is the gradient and $c$ is the $y$ -intercept. The equation must have $y$ on its own first: rearrange $2y = 6x + 4$ to $y = 3x + 2$ before reading off.

Gradient

= \dfrac{\text{rise}}{\text{run}} = \dfrac{3}{4}

and the line crosses the

y

-axis at 1, so

y = \frac{3}{4}x + 1

Example 1Find the gradient of the line through

(2, 7)

and

(6, -1)

1Change in

y

-1 - 7 = -8

2Change in

x

6 - 2 = 4

3Gradient

= \frac{-8}{4}

Answer

-2

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

Find the gradient of the line through $(-1, 3)$ and $(3, 11)$ .

Your answer

Show worked solution

Solution

1Change in $y$ : $11 - 3 = 8$ . Change in $x$ : $3 - (-1) = 4$ .
2Gradient $= \frac{8}{4} = 2$ .

Answer: $2$

Example 2Find the gradient and

y

-intercept of

4x + 2y = 10

1Make

y

the subject:

2y = -4x + 10

2Divide by 2:

y = -2x + 5

Answer

Gradient

-2

y

-intercept

5

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

What is the gradient of the line $3y - 6x = 12$ ?

Your answer

Show hint

Rearrange to $y = mx + c$ first.

Show worked solution

Solution

1 $3y = 6x + 12$ .
2 $y = 2x + 4$ , so the gradient is 2.

Answer: $2$

Tip — Subtract the coordinates in the same order on top and bottom: $(y_2 - y_1)$ over $(x_2 - x_1)$ .

Equation of a line through two points

Find the gradient $m$ from the two points. Substitute $m$ and one point into $y = mx + c$ and solve for $c$ .

Example 1Find the equation of the line through

(1, 4)

and

(3, 10)

1Gradient

= \frac{10 - 4}{3 - 1} = 3

2Substitute

(1, 4)

into

y = 3x + c

4 = 3 + c

, so

c = 1

Answer

y = 3x + 1

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

Find the equation of the line through $(2, 1)$ and $(5, 13)$ .

Your answer

Type it like y = 3x - 2.

Show worked solution

Solution

1Gradient $= \frac{13 - 1}{5 - 2} = 4$ .
2 $1 = 4(2) + c$ , so $c = -7$ .
3 $y = 4x - 7$

Answer: $y = 4x - 7$

Parallel and perpendicular lines

Parallel lines have the $\textbf{same gradient}$ . $y = 2x + 5$ and $y = 2x - 3$ are parallel.

Perpendicular lines (Higher) have gradients that multiply to $-1$ . If one gradient is $m$ , the other is $-\frac{1}{m}$ — the negative reciprocal. So $y = 2x + 5$ is perpendicular to $y = -\frac{1}{2}x + 1$ .

Example 1Find the equation of the line parallel to

y = -3x + 2

that passes through

(2, 5)

1Parallel, so the gradient is

-3

5 = -3(2) + c

, so

c = 11

Answer

y = -3x + 11

Example 2Find the equation of the line perpendicular to

y = \frac{2}{3}x - 1

through

(4, 1)

1The perpendicular gradient is

-\frac{3}{2}

1 = -\frac{3}{2}(4) + c

, so

1 = -6 + c

and

c = 7

Answer

y = -\frac{3}{2}x + 7

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

Find the equation of the line perpendicular to $y = 4x + 3$ that passes through $(8, 2)$ .

Your answer

Type it like y = -1/4x + 4 or y = -x/4 + 4.

Show worked solution

Solution

1Perpendicular gradient: $-\frac{1}{4}$ .
2 $2 = -\frac{1}{4}(8) + c$ , so $c = 4$ .

Answer: $y = -\frac{1}{4}x + 4$

Why negative reciprocals

Rotating a line through $90^\circ$ swaps its "rise" and "run" and turns one of them negative. A rise of 2 per run of 1 (gradient 2) becomes a rise of $-1$ per run of 2 (gradient $-\frac{1}{2}$ ). That is where the negative reciprocal comes from.

Think like an examiner

Always rearrange to $y = mx + c$ before stating a gradient — $2y = 6x + 4$ has gradient 3, not 6.
When plotting, a point that does not line up with the others is a signal to recheck its calculation.
Give equations in full: " $y = 3x + 1$ ", not " $m = 3$ , $c = 1$ ".

Straight lines

y = mx + c

m

= gradient,

c

y

-intercept.

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

Gradient from two points.

m_1 \times m_2 = -1

Perpendicular lines.

Watch out for these

Gradient of

(2, 7)

(6, -1)

\frac{6 - 2}{-1 - 7}

Change in

y

goes on top:

\frac{-1 - 7}{6 - 2} = -2

The gradient of

y = 5 - 2x

is 5.

Rewrite as

y = -2x + 5

: the gradient is

-2

The perpendicular gradient to 4 is

-4

It is the negative reciprocal:

-\frac{1}{4}

Thinking

y = 3

is a vertical line.

y = 3

is horizontal;

x = 3

is vertical.

Stretch yourself

The points $A(1, 2)$ , $B(5, 4)$ and $C(3, k)$ form a triangle with a right angle at $A$ . Find $k$ .

Hint — A right angle at $A$ means $AB$ and $AC$ are perpendicular.

Reveal solution

SolutionGradient of

AB = \frac{4 - 2}{5 - 1} = \frac{1}{2}

. So the gradient of

AC

must be

-2

\frac{k - 2}{3 - 1} = -2

, giving

k - 2 = -4

and

k = -2

\blacksquare

Questions students ask

What does a gradient of 0 mean?

The line is horizontal —

y

does not change as

x

increases. Its equation is

y = c

What is the gradient of a vertical line?

It is undefined, because the change in

x

is zero and you cannot divide by zero. Vertical lines have equations

x = a

Do I need to plot three points?

Two points define a line, but a third acts as a check. Examiners expect a table with several points for "draw the graph" questions.

Key takeaways

Substitute $x$ values to make a table, then plot and join with a ruler.
Gradient = change in $y$ ÷ change in $x$ .
In $y = mx + c$ , $m$ is the gradient and $c$ the $y$ -intercept.
Through two points: find $m$ , then substitute a point to find $c$ .
Parallel lines share a gradient; perpendicular gradients multiply to $-1$ .

Exam-style questions

Practise it the way it is examined

3 original questions · 10 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 43 marks

(a)Write down the gradient and the $y$ -intercept of the line $y = 7 - 4x$ .[2]
Show mark scheme (a)
- B1Gradient $-4$
- B1 $y$ -intercept $7$ (accept $(0, 7)$ )
(b)Write down the equation of a line parallel to $y = 7 - 4x$ .[1]
Show mark scheme (b)
- B1Any $y = -4x + c$ with $c \ne 7$

Question 2Grade 54 marks

A line passes through $(-2, 9)$ and $(4, -3)$ . Find its equation in the form $y = mx + c$ .[4]
Show mark scheme
- M1Gradient method: $\frac{-3 - 9}{4 - (-2)}$
- A1 $m = -2$
- M1Substitutes a point into $y = -2x + c$
- A1 $y = -2x + 5$
Answer: $y = -2x + 5$

Question 3Grade 73 marks

Line $L$ has equation $2y = x + 6$ . Find the equation of the line perpendicular to $L$ that passes through the point $(3, 1)$ .[3]
Show mark scheme
- M1Gradient of $L$ is $\frac{1}{2}$
- M1Perpendicular gradient $-2$ , substituted with $(3, 1)$ : $1 = -6 + c$
- A1 $y = -2x + 7$
Answer: $y = -2x + 7$

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.

0 / 0 first-time correct Streak 0Best at this level: 0

Generating a question…

Test yourself

Answer a few questions, then review anything you missed.

Prefer paper? Open the written self-check

Answer each question before opening its answer. This self-check does not record a quiz score.

What is the gradient of $y = 5x - 3$ ?
- $5$
- $-3$
- $3$
- $\frac{5}{3}$
Show answer and explanation
Answer: $5$
The coefficient of $x$ .
Where does $y = 2x + 7$ cross the $y$ -axis?
- $(0, 7)$
- $(7, 0)$
- $(0, 2)$
- $(-3.5, 0)$
Show answer and explanation
Answer: $(0, 7)$
The intercept $c = 7$ .
Which point lies on $y = 4x - 1$ ?
- $(2, 7)$
- $(7, 2)$
- $(1, 5)$
- $(0, 1)$
Show answer and explanation
Answer: $(2, 7)$
$4(2) - 1 = 7$ .
Gradient of the line through $(1, 2)$ and $(4, 11)$ ?
- $3$
- $\frac{1}{3}$
- $9$
- $-3$
Show answer and explanation
Answer: $3$
$\frac{9}{3} = 3$ .
Gradient of $3y = 12x - 6$ ?
- $4$
- $12$
- $-2$
- $3$
Show answer and explanation
Answer: $4$
$y = 4x - 2$ .
Which line is parallel to $y = -x + 4$ ?
- $y = 3 - x$
- $y = x + 4$
- $y = -4x$
- $y = 4$
Show answer and explanation
Answer: $y = 3 - x$
Both have gradient $-1$ .
A line perpendicular to $y = -\frac{1}{3}x + 2$ has gradient…
- $3$
- $-3$
- $\frac{1}{3}$
- $-\frac{1}{3}$
Show answer and explanation
Answer: $3$
$-\frac{1}{3} \times 3 = -1$ .
Equation of the line through $(0, -2)$ and $(4, 6)$ ?
- $y = 2x - 2$
- $y = 2x + 6$
- $y = \frac{1}{2}x - 2$
- $y = 4x - 2$
Show answer and explanation
Answer: $y = 2x - 2$
Gradient $\frac{8}{4} = 2$ , intercept $-2$ .

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