What is A9: Straight-line graphs?

A non-vertical straight line has equation y=mx+c, where m is the gradient and (0,c) is the y-intercept. Plot a line by calculating at least two accurate coordinate pairs and joining them carefully with a ruler.

How many marks are A9 questions worth?

In verified past-paper appearances, A9 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has A9 appeared in?

A9 has appeared in 27 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

Plot graphs of straight-line equations; use y = mx + c to identify parallel and perpendicular lines; find the equation of a line through two given points, or one point with a given gradient

Straight-line graphs

Worked answers, methods and verified real exam appearances for A9 on Edexcel GCSE Maths 1MA1.

Explanation

A non-vertical straight line has equation $y=mx+c$ , where $m$ is the gradient and $(0,c)$ is the $y$ -intercept. Plot a line by calculating at least two accurate coordinate pairs and joining them carefully with a ruler.
From two points, find $m=\dfrac{y_2-y_1}{x_2-x_1}$ , then substitute either point to find $c$ .
Parallel lines have equal gradients.
Higher tier: perpendicular non-vertical lines have gradients whose product is $-1$ .
Examiners expect the gradient calculation, substitution for the intercept and a final equation in a requested form.

A straight line

y=mx+c

crosses the vertical axis at

(0,c)

and has constant gradient.

Worked example

Find the equation of the line through $(-2,5)$ and $(4,-1)$ .

1. $m=\dfrac{-1-5}{4-(-2)}=\dfrac{-6}{6}=-1$ .
2.Use $y=-x+c$ and substitute $(-2,5)$ : $5=2+c$ .
3. $c=3$ , so the equation is $y=-x+3$ .

Answer: $y=-x+3$ .

Common mistakes

Don't subtract coordinates in different orders in the gradient numerator and denominator.
Don't use the $x$ -intercept as $c$ in $y=mx+c$ .
Don't change only the sign of a gradient to make a perpendicular line (Higher tier).

Exam tip

For a line through two points, show the gradient first and then substitute one point to find $c$ .

Worked practice

Tier 1 · Easy

Write the equation of the line with gradient

4

and

y

-intercept

-7

(1)

(Total for Question 1 is 1 mark)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$y=4x-7$	1	In $y=mx+c$ , use $m=4$ and $c=-7$ . This gives $y=4x-7$ .

Tier 2 · Standard

Line

A

has equation

4y=6x+7

. Line

B

passes through the points

(-2,4)

and

(2,10)

. Show that the two lines are parallel.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	Each line has gradient $\dfrac{3}{2}$ , and their $y$ -intercepts differ ( $\tfrac74$ and $7$ ), so the lines are parallel and distinct.	3	Rearranging line $A$ gives $y=\frac32x+\frac74$ , so its gradient is $\frac32$ and its $y$ -intercept is $\frac74$ . The gradient of line $B$ is $(10-4)/(2-(-2))=6/4=\frac32$ , and substituting $(2,10)$ gives $y=\frac32x+7$ . The gradients are equal, so the lines never converge; the $y$ -intercepts differ, so the lines are not the same line. Hence they are parallel.

Tier 3 · Hard

Find an equation of the line through

(-6,5)

that is perpendicular to

7x+4y=13

. Give your answer in the form

y=mx+c

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$y=\frac47x+\frac{59}{7}$	4	Rearrange the given line: $4y=13-7x$ , so $y=-\frac74x+\frac{13}{4}$ and its gradient is $-\frac74$ . A perpendicular line has gradient $\frac47$ . Write $y=\frac47x+c$ and substitute $(-6,5)$ : $5=-\frac{24}{7}+c$ , so $c=\frac{59}{7}$ . Hence $y=\frac47x+\frac{59}{7}$ .

Tier 1 · Easy

The point

P

lies on the line

y=3x-5

and has

x

-coordinate

-2

. Find the coordinates of

P

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$P=(-2,-11)$	2	Substitute $x=-2$ into the line equation: $y=3(-2)-5=-6-5=-11$ . Therefore $P=(-2,-11)$ .

Tier 2 · Standard

Find an equation of the line with gradient

-2

that passes through

(4,1)

. Give your answer in the form

y=mx+c

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$y=-2x+9$	3	Write $y=-2x+c$ and substitute $(4,1)$ : $1=-2(4)+c=-8+c$ . Hence $c=9$ , so the equation is $y=-2x+9$ .

Tier 3 · Hard

The line through

(-4,7)

and

(2,-5)

also passes through

(5,k)

. Work out the value of

k

(3)

(Total for Question 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$k=-11$	3	The gradient between the first two points is $\dfrac{-5-7}{2-(-4)}=\dfrac{-12}{6}=-2$ . From $x=2$ to $x=5$ is an increase of $3$ , so $y$ changes by $-2\times3=-6$ . Hence $k=-5-6=-11$ .

Tier 2 · Standard

Line

L

is parallel to the line

3y=6x-12

and passes through

(-1,5)

. Find an equation of

L

in the form

y=mx+c

(3)

(Total for Question 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$y=2x+7$	3	Rearrange the given line to $y=2x-4$ , so a parallel line has gradient $2$ . Write $L$ as $y=2x+c$ and substitute $(-1,5)$ : $5=-2+c$ , so $c=7$ . Therefore $L$ is $y=2x+7$ .

Tier 3 · Hard

The line

px+2y=8

passes through

(4,-2)

. Work out the value of

p

. Hence find the gradient, the

x

-intercept and the

y

-intercept of the line.

(4)

(Total for Question 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$p=3$ Gradient $-\dfrac32$ $x$ -intercept $\dfrac83$ $y$ -intercept $4$	4	Substitute $(4,-2)$ : $4p+2(-2)=8$ , so $4p=12$ and $p=3$ . The line is $3x+2y=8$ , or $y=-\dfrac32x+4$ , giving gradient $-\dfrac32$ and $y$ -intercept $4$ . Setting $y=0$ gives $3x=8$ , so the $x$ -intercept is $\dfrac83$ .

Tier 3 · Hard

The points are

A(-3,7)

B(1,-1)

and

C(4,-7)

. Show that the three points are collinear. Find an equation of the line through them.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	The gradients $AB$ and $BC$ are both $-2$ $y=-2x+1$	4	The gradient of $AB$ is $\dfrac{-1-7}{1-(-3)}=-2$ . The gradient of $BC$ is $\dfrac{-7-(-1)}{4-1}=-2$ . Equal gradients through the common point $B$ show the three points are collinear. Write $y=-2x+c$ and substitute $B(1,-1)$ : $-1=-2+c$ , so $c=1$ and the equation is $y=-2x+1$ .

Q10

Tier 3 · Hard

Higher only: The endpoints of a line segment are

A(-2,1)

and

B(4,5)

. Find the equation of the perpendicular bisector of

AB

in the form

y=mx+c

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$y=-\dfrac32x+\dfrac92$	5	The midpoint is $\left(\dfrac{-2+4}{2},\dfrac{1+5}{2}\right)=(1,3)$ . The gradient of $AB$ is $\dfrac{5-1}{4-(-2)}=\dfrac23$ , so the perpendicular gradient is $-\dfrac32$ . Using the unique midpoint gives $y-3=-\dfrac32(x-1)$ , hence $y=-\dfrac32x+\dfrac92$ . A line segment has one midpoint and one perpendicular direction, so the configuration is unique.

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-11	2H	Q9	2	Allowed	Higher	QP MS
2019-06	2H	Q16	3	Allowed	Higher	QP MS
2022-06	1H	Q20	4	Non-calculator	Higher	QP MS
2023-11	1H	Q12	3	Non-calculator	Higher	QP MS
2021-11	2H	Q11	5	Allowed	Higher	QP MS
2023-11	3F	Q28	3	Allowed	Foundation	QP MS
2024-06	1H	Q23	5	Non-calculator	Higher	QP MS
2022-11	3F	Q9	3	Allowed	Foundation	QP MS
2019-06	2H	Q2	3	Allowed	Higher	QP MS
2021-11	1F	Q18	3	Non-calculator	Foundation	QP MS
2024-11	2H	Q21	5	Allowed	Higher	QP MS
2024-11	3F	Q27	3	Allowed	Foundation	QP MS
2019-11	2H	Q25	5	Allowed	Higher	QP MS
2022-06	2F	Q17	3	Allowed	Foundation	QP MS
2023-11	3H	Q9	3	Allowed	Higher	QP MS
2023-06	1F	Q9	4	Non-calculator	Foundation	QP MS
2023-06	1H	Q15	3	Non-calculator	Higher	QP MS
2019-11	3F	Q17	4	Allowed	Foundation	QP MS
2024-06	3F	Q19	3	Allowed	Foundation	QP MS
2024-06	1F	Q25	4	Non-calculator	Foundation	QP MS
2023-06	3H	Q23	4	Allowed	Higher	QP MS
2023-11	1F	Q14	3	Non-calculator	Foundation	QP MS
2024-11	1H	Q20	4	Non-calculator	Higher	QP MS
2022-06	2H	Q12	2	Allowed	Higher	QP MS
2024-11	3H	Q16	5	Allowed	Higher	QP MS
2024-06	1H	Q6	4	Non-calculator	Higher	QP MS
2019-06	2F	Q21	3	Allowed	Foundation	QP MS

Other points in A Algebra · equations and reasoning

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