What is 3.1: Straight lines?

A non-vertical straight line has constant gradient m and may be written as y-y_1=m(x-x_1) through a known point or rearranged into ax+by+c=0.

How many marks are 3.1 questions worth?

3.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 3.1 appeared in?

We have not yet indexed a verified past-paper appearance for 3.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

3.1

Understand and use the equation of a straight line, including the forms y − y1 = m(x − x1) and ax + by + c = 0; gradient conditions for two straight lines to be parallel or perpendicular; use straight line models in a variety of contexts.

Draft — not yet indexed

Straight lines

Worked answers and methods for 3.1 on Edexcel A-level Maths 9MA0.

Explanation

A non-vertical straight line has constant gradient $m$ and may be written as $y-y_1=m(x-x_1)$ through a known point or rearranged into $ax+by+c=0$ .
Parallel lines have equal gradients, while non-vertical perpendicular lines have gradients whose product is $-1$ .
For two known points, first calculate $m=(y_2-y_1)/(x_2-x_1)$ , then substitute one point into a line equation and check the other point.
In a linear model, interpret the gradient and intercept in context and respect practical restrictions.
A common error is to round a limiting input upwards when this breaks the stated constraint.

Worked example

The line $L$ has equation $2x-5y+7=0$ . Find an equation of the line through $(4,-1)$ that is perpendicular to $L$ . Write your equation as $ax+by+c=0$ with integer coefficients.

1.Rearrange $L$ to $y=\dfrac{2}{5}x+\dfrac{7}{5}$ , so its gradient is $\dfrac{2}{5}$ .
2.A perpendicular line has gradient $-\dfrac{5}{2}$ .
3.Hence $y+1=-\dfrac{5}{2}(x-4)$ , which rearranges to $5x+2y-18=0$ .

Answer: $5x+2y-18=0$

Common mistakes

Don't treat the constant $c$ in $ax+by+c=0$ as the $y$ -intercept without first rearranging the equation.
Don't use the negative reciprocal rule on the coefficients without first finding the line's gradient.

Exam tip

When asked for a perpendicular line, extract the original gradient, use its negative reciprocal, then substitute the given point.

Worked practice

Tier 1 · Easy

Find the equation of the line with gradient

-3

that passes through

(2,4)

. Write it as

y=mx+c

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$y=-3x+10$	2
Notes
Use point-gradient form: $y-4=-3(x-2)$ . Expanding gives $y-4=-3x+6$ , so $y=-3x+10$ .

(2 marks)

Tier 2 · Standard

The lines

y=2x+1

and

y=-x+7

meet at

P

. Find the coordinates of

P

and an equation of the line through

P

that is parallel to

3x+y=4

. Write your equation as

ax+by+c=0

. Use integer coefficients with greatest common divisor

1

, and choose

a>0

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$P=(2,5)$ $3x+y-11=0$	4
Notes
At $P$ , $2x+1=-x+7$ , so $3x=6$ and $x=2$ ; then $y=5$ . The line $3x+y=4$ has gradient $-3$ , so a parallel line through $(2,5)$ is $y-5=-3(x-2)$ . Rearranging gives $3x+y-11=0$ .

(4 marks)

Tier 3 · Hard

A delivery company models the charge

C

pounds for a journey of

d

kilometres by a straight line. Journeys of

8

km and

14

km cost

\pounds 19.60

and

\pounds 29.80

respectively. Find the model for

C

in terms of

d

, interpret its gradient, and find the greatest whole number of kilometres that can be travelled for at most

\pounds 45

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$C=1.7d+6$ The model charges $\pounds 1.70$ per kilometre. $22$ km	5
Notes
The gradient is $(29.80-19.60)/(14-8)=10.20/6=1.70$ , representing the charge per kilometre. Write $C=1.7d+c$ and use $(8,19.60)$ : $19.60=1.7(8)+c$ , so $c=6$ . For the budget, solve $1.7d+6\le45$ , giving $d\le390/17\approx22.94$ . The greatest permitted whole number is therefore $22$ km; $23$ km would cost $\pounds 45.10$ .

(5 marks)

Tier 1 · Easy

The points

A(-2,7)

and

B(4,-5)

lie on a straight line. Find an equation of

AB

in the form

ax+by+c=0

. Use integer coefficients with greatest common divisor

1

, and choose

a>0

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$2x+y-3=0$	3
Notes
The gradient of $AB$ is $(-5-7)/(4-(-2))=-12/6=-2$ . Using point $A$ , $y-7=-2(x+2)$ , so $y=-2x+3$ . Rearranging gives $2x+y-3=0$ .

(3 marks)

Tier 2 · Standard

The lines

L_1

and

L_2

have equations

y=(k-1)x

and

2x+(k+3)y=34

respectively, where

k

is a constant. Given that

L_1

and

L_2

are perpendicular, find

k

and the coordinates of their point of intersection.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$k=5$ $(1,4)$	4
Notes
If $k=-3$ , $L_2$ is vertical but $L_1$ is not horizontal, so the lines are not perpendicular. Otherwise their gradients are $k-1$ and $-2/(k+3)$ . Perpendicularity gives $(k-1)[-2/(k+3)]=-1$ , so $2(k-1)=k+3$ and $k=5$ . The lines are then $y=4x$ and $2x+8y=34$ . Substitution gives $34x=34$ , so $x=1$ and $y=4$ .

(4 marks)

Tier 3 · Hard

The points

A(2,-1)

and

B(p,7)

lie on a line that is perpendicular to

3x-2y+5=0

. Find

p

and the equation of

AB

. The line

AB

meets the positive coordinate axes at

P

and

Q

. Find the exact area of triangle

OPQ

, where

O

is the origin.

(6)

(Total for Question 6 is 6 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$p=-10$ $AB: 2x+3y-1=0$ Area $=\dfrac{1}{12}$ square units	6
Notes
The line $3x-2y+5=0$ has gradient $3/2$ , so $AB$ has gradient $-2/3$ . Also the gradient from $A$ to $B$ is $8/(p-2)$ , hence $8/(p-2)=-2/3$ and $p=-10$ . Through $A$ , $y+1=-\dfrac{2}{3}(x-2)$ , which gives $2x+3y-1=0$ . Its positive intercepts are $P=(1/2,0)$ and $Q=(0,1/3)$ , so the area is $\tfrac12(1/2)(1/3)=1/12$ square units.

(6 marks)

Tier 2 · Standard

The perpendicular bisector of the segment joining

A(-3,2)

B(5,6)

is the line

L

. Find an equation of

L

in the form

ax+by+c=0

. Use integer coefficients with greatest common divisor

1

, and choose

a>0

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$2x+y-6=0$	4
Notes
The midpoint of $AB$ is $(1,4)$ and the gradient of $AB$ is $(6-2)/[5-(-3)]=1/2$ . Therefore the perpendicular bisector has gradient $-2$ . Its equation is $y-4=-2(x-1)$ , which rearranges to $2x+y-6=0$ .

(4 marks)

Tier 3 · Hard

The point

P(7,-2)

does not lie on the line

L

with equation

3x-4y+12=0

. The point

Q

L

is closest to

P

. Using straight-line methods, find the exact coordinates of

Q

and the distance

PQ

. The reflection of

P

L

is the point

R

. Find the exact coordinates of

R

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$Q=\left(\dfrac{52}{25},\dfrac{114}{25}\right)$ $PQ=\dfrac{41}{5}$ $R=\left(-\dfrac{71}{25},\dfrac{278}{25}\right)$	6
Notes
The gradient of $L$ is $3/4$ , so the perpendicular through $P$ has equation $y+2=-\dfrac43(x-7)$ , or $4x+3y-22=0$ . Solving this with $3x-4y+12=0$ gives $Q=(52/25,114/25)$ . Hence $PQ=\sqrt{(123/25)^2+(164/25)^2}=41/5$ . Since reflection in $L$ makes $Q$ the midpoint of $PR$ , $R=2Q-P=(-71/25,278/25)$ .

(6 marks)

Tier 3 · Hard

The line

L_k

has equation

(k+1)x+(2-k)y=3k+4

, where

k

is a real constant. Show that every line in this family passes through a fixed point

P

. Find

P

. Find the value of

k

for which

L_k

is parallel to the line

2x-3y=7

. For this value of

k

, find the exact area of the triangle formed by

L_k

and the coordinate axes.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$P=\left(\dfrac{10}{3},\dfrac{1}{3}\right)$ $k=-7$ Area $=\dfrac{289}{108}$ square units	6
Notes
Rearrange the family as $k(x-y-3)+(x+2y-4)=0$ . A point on every member must satisfy $x-y-3=0$ and $x+2y-4=0$ , giving $P=(10/3,1/3)$ . Parallel lines have proportional $x$ - and $y$ -coefficients, so $(k+1)/(2-k)=2/(-3)$ . Hence $-3(k+1)=2(2-k)$ and $k=-7$ . The selected line is $6x-9y=17$ , with intercepts $(17/6,0)$ and $(0,-17/9)$ . Its triangle with the coordinate axes therefore has area $\tfrac12(17/6)(17/9)=289/108$ square units.

(6 marks)

Q10

Tier 3 · Hard

10.

The points

A(-3,1)

B(5,3)

and

C(k,9)

are consecutive vertices of a parallelogram

ABCD

. The midpoint of its diagonals lies on the line

2x-y=1

. Using straight-line methods, find

k

and the coordinates of

D

. Find an equation of the diagonal

BD

and the exact perpendicular distance from

A

BD

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$k=9$ $D=(1,7)$ $BD: x+y-8=0$ Distance $=5\sqrt2$	6
Notes
The diagonals of a parallelogram bisect each other, so their midpoint is the midpoint of $AC$ , namely $M=((k-3)/2,5)$ . Since $M$ lies on $2x-y=1$ , $k-3-5=1$ , giving $k=9$ . Also $A+C=B+D$ , so $D=A+C-B=(1,7)$ . The gradient of $BD$ is $(7-3)/(1-5)=-1$ , hence $BD$ has equation $y-3=-(x-5)$ , or $x+y-8=0$ . The perpendicular distance from $A(-3,1)$ to this line is $\|-3+1-8\|/\sqrt2=10/\sqrt2=5\sqrt2$ .

(6 marks)

Verified exam appearances

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Other points in 3 Coordinate geometry in the (x, y) plane

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