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 $y = m x + c$ . 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 $y = 2 x + 1$ 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.

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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.

x = - 1

y = - 5

x = 0

y = - 2

x = 1

y = 1

x = 2

y = 4

AnswerPoints

(- 1, - 5)

(0, - 2)

(1, 1)

(2, 4)

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

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

Your answer

Gradient and intercept

Gradient $= \frac{change in y}{change in x}$ , 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 $y = m x + c$ , $m$ is the gradient and $c$ is the $y$ -intercept. The equation must have $y$ on its own first: rearrange $2 y = 6 x + 4$ to $y = 3 x + 2$ before reading off.

Gradient

= \frac{rise}{run} = \frac{3}{4}

and the line crosses the

y

-axis at 1, so

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

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

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

Your answer

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 $y = m x + c$ and solve for $c$ .

1Gradient

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

2Substitute

(1, 4)

into

y = 3 x + c

4 = 3 + c

, so

c = 1

Answer

y = 3 x + 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.

Parallel and perpendicular lines

Parallel lines have the $same gradient$ . $y = 2 x + 5$ and $y = 2 x - 3$ 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 $y = 2 x + 5$ is perpendicular to $y = - \frac{1}{2} x + 1$ .

1Parallel, so the gradient is

- 3

5 = - 3 (2) + c

, so

c = 11

Answer

y = - 3 x + 11

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

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

Your answer

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

Why negative reciprocals

Rotating a line through $9 0^{\circ}$ 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 $y = m x + c$ before stating a gradient — $2 y = 6 x + 4$ 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: " $y = 3 x + 1$ ", not " $m = 3$ , $c = 1$ ".

Straight lines

y = m x + c

m

= gradient,

c

y

-intercept.

m = \frac{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 - 2 x

is 5.

Rewrite as

y = - 2 x + 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 $A B$ and $A C$ are perpendicular.

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 $y = m x + c$ , $m$ 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 $y = 7 - 4 x$ .[2]
(b)Write down the equation of a line parallel to $y = 7 - 4 x$ .[1]

Question 2Grade 54 marks

A line passes through $(- 2, 9)$ and $(4, - 3)$ . Find its equation in the form $y = m x + c$ .[4]

Question 3Grade 73 marks

Line $L$ has equation $2 y = x + 6$ . Find the equation of the line perpendicular to $L$ that passes through the point $(3, 1)$ .[3]

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Straight Line Graphs

Plotting from a table

Gradient and intercept

Equation of a line through two points

Parallel and perpendicular lines

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Related topics

Straight Line Graphs

Plotting from a table

Gradient and intercept

Equation of a line through two points

Parallel and perpendicular lines

Practise it the way it is examined

Skill drill

Practise this topic

Related topics