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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

LearnYMA01WMA11 · Pure Mathematics 1Coordinate geometry in the (x, y) planeGradient, length, midpoint and equations of lines

Coordinate geometry in the (x, y) plane · Subtopic 1 of 2

Gradient, Length, Midpoint and Equations of Lines

The three measurements of a line segment (gradient, length and midpoint), the three forms of the equation of a straight line, and how to find where lines cross.

Cambridge 9709 Paper 1Edexcel YMA01 WMA11Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 35 minRevision 6 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Find the gradient, length and midpoint of a line segment, and work backwards from them.
Write the equation of a line in the forms $y=mx+c$ , $y-y_1=m(x-x_1)$ and $ax+by+c=0$ .
Read the gradient and intercept from any form of the equation.
Find where two lines intersect, check whether points are collinear, and find simple areas.

Pause and think

A line goes through $(0,1)$ and $(2,7)$ . Without any formula: how much does $y$ go up each time $x$ goes up by $1$ ?

Section 2 of 14

Core idea

Start hereDescribing a road with two numbers

A road sign says "Gradient 1 in 10": for every $10$ m you go along, you climb $1$ m. A map gives the road's start and end as grid references. From just two points you can work out how steep the road is, how long it is, where its halfway café should go, and a formula for every point on it. That's coordinate geometry: turning shapes into numbers so algebra can answer geometric questions.

The big idea

From two points $A(x_1,y_1)$ and $B(x_2,y_2)$ you get the gradient $m=\dfrac{y_2-y_1}{x_2-x_1}$ , the length $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ , the midpoint $\left(\frac{x_1+x_2}2,\frac{y_1+y_2}2\right)$ , and the line's equation $y-y_1=m(x-x_1)$ .

1. Gradient: rise over run

Gradient

m=\frac{\text{rise}}{\text{run}}=\frac{y_2-y_1}{x_2-x_1}

$(x_1,y_1)$ , $(x_2,y_2)$: the two points; the small numbers are labels, not powers.
$y_2-y_1$: the rise: how far up (negative means down).
$x_2-x_1$: the run: how far across.

A(1, 2) to B(5, 10): the run and the rise

Gradient

=\dfrac{\text{rise}}{\text{run}}=\dfrac84=2

. Length

AB=\sqrt{4^2+8^2}=\sqrt{80}=4\sqrt5

. Midpoint

=\left(\dfrac{1+5}{2},\dfrac{2+10}{2}\right)=(3,6)

Worked exampleGradients of two segments

Find the gradient of $AB$ with $A(1,2)$ , $B(5,10)$ , and of $PQ$ with $P(-2,3)$ , $Q(4,-1)$ .

Goal: Two gradients, with correct signs.
Strategy: Subtract in the same order on top and bottom; use brackets for negatives.

$m_{AB}=\frac{10-2}{5-1}=\frac84=2$
Up 2 for every 1 across.
$m_{PQ}=\frac{-1-3}{4-(-2)}=\frac{-4}{6}=-\frac23$
$4-(-2)=6$ ; negative gradient: downhill.

Answer

2

and

-\frac23

✓ CheckReversing the order:

\frac{3-(-1)}{-2-4}=\frac4{-6}=-\frac23

✓ (order doesn't matter if consistent).

$m>0$ : uphill (left to right); $m<0$ : downhill.
$m=0$ : horizontal line $y=k$ .
Vertical line $x=k$ : the run is $0$ , so the gradient is undefined.

Common misconceptionMixed order

\frac{y_2-y_1}{x_1-x_2}

gives the wrong sign. Same point first on top and bottom.

2. Length: Pythagoras on the grid

The distance formula

$\text{run}=x_2-x_1,\qquad\text{rise}=y_2-y_1$
The two shorter sides of a right-angled triangle with hypotenuse $AB$ .
$AB^2=(x_2-x_1)^2+(y_2-y_1)^2$
Pythagoras: hypotenuse² = sum of the other two squared.
$AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
Square root (length is positive).

For

A(1,2)

B(5,10)

\sqrt{16+64}=\sqrt{80}=4\sqrt5\approx8.94

3. Midpoint: the average

Halfway along, you've done half the run and half the rise: $x=x_1+\frac12(x_2-x_1)=\frac{x_1+x_2}2$ , and similarly for $y$ . So the midpoint's coordinates are the averages.

Worked exampleWorking backwards from a midpoint

$M(4,1)$ is the midpoint of $A(1,-3)$ and $B$ . Find $B$ .

Goal: Coordinates of $B(x,y)$ .
Strategy: Write the midpoint formula with unknowns and solve each coordinate.

$\frac{1+x}2=4\Rightarrow1+x=8\Rightarrow x=7$
$x$ -coordinates.
$\frac{-3+y}2=1\Rightarrow-3+y=2\Rightarrow y=5$
$y$ -coordinates.

Answer

B(7,5)

✓ CheckMidpoint of

(1,-3)

and

(7,5)

(4,1)

✓.

4. The equation of a line

Point–gradient form

y-y_1=m(x-x_1)

$\text{Let }(x,y)\text{ be any other point on the line.}$
We want a rule every point obeys.
$\frac{y-y_1}{x-x_1}=m$
The gradient from $(x_1,y_1)$ to $(x,y)$ must equal $m$ (a straight line has constant gradient).
$y-y_1=m(x-x_1)$
Multiply both sides by $(x-x_1)$ .

Expanding gives

y=mx+c

, where

c=y_1-mx_1

is the

y

-intercept.

y = mx + 1 as the gradient m changes

m = -2.0

The gradient is how much

y

goes up when

x

goes up by

1

Worked exampleLine through $(3,-1)$ with gradient $-\tfrac23$

Find the equation in the form $ax+by+c=0$ with integers.

Goal: Integer coefficients.
Strategy: Point–gradient form, clear the fraction, collect.

$y+1=-\tfrac23(x-3)$
$y-(-1)=y+1$ .
$3y+3=-2(x-3)$
Multiply both sides by $3$ .
$3y+3=-2x+6$
Expand.
$2x+3y-3=0$
Add $2x$ , subtract $6$ .

Answer

2x+3y-3=0

✓ Check

(3,-1)

6-3-3=0

✓. Gradient:

y=-\frac23x+1

✓.

Worked exampleWhere two lines meet

Find where $y=2x$ meets $x+y=9$ .

Goal: The intersection point.
Strategy: The point satisfies both equations: solve simultaneously.

$x+2x=9\Rightarrow x=3$
Substitute $y=2x$ .
$y=6$
Back into $y=2x$ .

Answer

(3,6)

✓ Check

3+6=9

✓.

Pause and think

Where does $3x+4y=24$ cross each axis, and how far apart are those points?

Summary

Gradient = rise/run.
Length by Pythagoras; midpoint by averages.
$y-y_1=m(x-x_1)$ ; $y=mx+c$ .
Intersections: solve simultaneously.

Gradient: Rise per unit run; steepness.
Intercept: Where a line crosses an axis.

HistoryWhy they are called Cartesian coordinates. In 1637 the French philosopher René Descartes published La Géométrie, showing how to describe curves with equations. Pierre de Fermat had the same idea at about the same time. Coordinates are named "Cartesian" after Descartes, whose Latin name was Renatus Cartesius.

Fun factThe fly on the ceiling (probably a legend). A popular story says Descartes invented coordinates while lying in bed watching a fly crawl across the ceiling, realising he could describe its position by its distances from two walls. There is no evidence it is true, but it is a perfect picture of what

(x,y)

means.

Section 3 of 14

Key formulae

For A(x₁, y₁) and B(x₂, y₂)

\begin{gathered}m=\frac{y_2-y_1}{x_2-x_1}\qquad AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \\[4pt] \text{midpoint }M=\left(\frac{x_1+x_2}2,\ \frac{y_1+y_2}2\right)\end{gathered}

Three forms of a straight line

\begin{gathered}y=mx+c\qquad y-y_1=m(x-x_1)\qquad ax+by+c=0 \\[3pt] \text{from }ax+by+c=0:\ \ m=-\frac ab\end{gathered}

NoteWhich form to give? Use

y-y_1=m(x-x_1)

to find the line: it needs no rearranging. Then give the answer in the form the question asks for. "

ax+by+c=0

where

a

b

and

c

are integers" means clear all fractions first.

Section 4 of 14

Worked examples

Example 1 · All three measurements0 / 3 steps

A

(-3,4)

and

B

(5,-2)

. Find the gradient, the length and the midpoint of

AB

Example 2 · Equation from two points0 / 3 steps

Find the equation of the line through

(1,2)

and

(4,11)

in the form

ax+by+c=0

Example 3 · Working backwards from a midpoint0 / 2 steps

M(3,5)

is the midpoint of

AB

, where

A

(1,2)

. Find

B

Example 4 · Where two lines cross0 / 2 steps

Find the point of intersection of

2x+y=7

and

x-y=2

Section 5 of 14

Explore the graph

Two points, three measurements

Loading the interactive graph…

Try this

The two value rows show the gradient and the length of the segment. Check they match Example 1: $-\tfrac34$ and $10$ .
Make $b=d$ . What is the gradient of a horizontal line?
Make $a=c$ . Why does the gradient row show an error? (A vertical line has an undefined gradient, and its equation is $x=a$ .)

The middle point is the midpoint. Try to predict where it will be before you move a slider.

TrickThe step trick for midpoints and ends. Find the step from one point to the midpoint (for example, right

2

, up

3

), then take the same step again to get the other end. It is quicker than solving two equations and much harder to get wrong.

In the real worldScreens count downwards. On a computer screen, pixel coordinates start at the top-left corner and

y

increases downwards. Game designers flip the sign of every gradient, which is why a "rising" line on screen has a negative gradient in their code.

Section 6 of 14

Common mistakes

IncorrectGradient

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

Correct

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

Gradient is "up over across": change in

y

on top.

Incorrect

3x+4y=12

has gradient

3

Correct

-\tfrac34

Rearrange to

y=-\tfrac34x+3

before reading the gradient.

Incorrect

\sqrt{(5-(-3))^2+(-2-4)^2}=\sqrt{2^2+6^2}

Correct

\sqrt{8^2+6^2}

5-(-3)=8

: subtracting a negative adds.

Incorrect

y-2=3(x-1)\Rightarrow y=3x-3-2

Correct

y=3x-3+2=3x-1

Move the

-2

across by adding

2

Gradient is run over rise
Gradient is rise over run
Gradient is $\dfrac{y_2-y_1}{x_2-x_1}$ and I keep the order the same on top and bottom

Section 7 of 14

Exam tip

Exam tipKeep fractions exact. If a gradient is

-\tfrac34

, keep it as a fraction. Decimals like

-0.75

are fine, but

\tfrac13

0.333

loses accuracy marks in the next step. Answers such as

\sqrt{52}

should be simplified to

2\sqrt{13}

when asked for exact form.

Section 8 of 14

Guided practice

SA-9709-P1-CG-G014 marksL1 FoundationCalculatorEquations

A

(1,\,2)

and

B

(5,\,10)

. Find the gradient of

AB

, the midpoint of

AB

, and the length of

AB

in exact form.

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P1-CG1-0012 marksL1 FoundationCalculatorGradient & midpoint

Find the gradient of the line through

(-3,4)

and

(5,-2)

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG1-0032 marksL2 StandardCalculatorGradient & midpoint

M(3,5)

is the midpoint of

AB

, where

A

(1,2)

. Find the coordinates of

B

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG-0012 marksL1 FoundationCalculatorEquations

Find the equation of the line with gradient

3

passing through

(2,\,-1)

, in the form

y=mx+c

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG-0032 marksL1 FoundationCalculatorEquations

The line

L

has equation

3x+4y=12

. State the gradient of

L

and the

y

-coordinate where

L

crosses the

y

-axis.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG1-0062 marksL2 StandardCalculatorEquations

The line

y=\tfrac23x-4

can be written as

2x+by+c=0

, where

b

and

c

are integers. Find

b

and

c

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG-0043 marksL2 StandardCalculatorEquations

Find the point of intersection of the lines

y=2x-1

and

x+3y=11

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG1-0073 marksL2 StandardCalculatorIntersections

The points

(1,3)

(3,7)

and

(6,k)

lie on a straight line. Find

k

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P1-CG-0143 marksL2 StandardCalculatorEquations

The distance between

P(k,\,3)

and

Q(4,\,-1)

5

. Find the possible values of

k

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P1-CG1-0105 marksL5 ExtensionCalculatorGradient & midpoint

A

(0,2)

and

B

(6,5)

. The point

C

lies on the

x

-axis and is the same distance from

A

as from

B

. Find the

x

-coordinate of

C

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Gradient: change in $y$ over change in $x$ . Length: Pythagoras. Midpoint: average the coordinates.
Find the line with $y-y_1=m(x-x_1)$ , then rearrange to the form asked for.
For $ax+by+c=0$ , the gradient is $-\tfrac ab$ .
Intersections: solve the two equations simultaneously.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P1-CG1-Q011 markL1 FoundationCalculatorEquations

The gradient of the line

y=5-3x

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P1-CG1-Q021 markL1 FoundationCalculatorEquations

The line through

(2,-4)

parallel to the

x

-axis has equation

Salam Academy Original Practice · Proposed mark scheme

Q3 · SA-9709-P1-CG1-Q031 markL1 FoundationCalculatorGradient & midpoint

The midpoint of

(x_1,y_1)

and

(x_2,y_2)

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Mathematics
Where: WMA11 (Pure 1) · Coordinate geometry in the (x, y) plane
Expected: Equation of a straight line in the forms y − y₁ = m(x − x₁) and ax + by + c = 0, with gradient, length, midpoint and intersections (WMA11).
Calculator: WMA11 (straight lines) is non-calculator; WMA12 (circles) allows a calculator.
Typical style: Short structured parts, often with a sketch given to label.
Open in: YMA01 · WMA11 9MA0 · Paper 1 9MA0 · Paper 2

Section 14 of 14

Past paper questions

3 questions (22 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

WMA11/01 · June 2025Question 57 marks
A perpendicular line, where two lines meet, and a distance to the origin.
Question paper, page 10 Mark scheme
WMA11/01 · June 2024Question 99 marks
Line through two points, a point at a given distance and a triangle area.
Question paper, page 22 Mark scheme
WMA11/01 · June 2023Question 76 marks
A region bounded by lines, described with inequalities.
Question paper, page 18 Mark scheme

All YMA01 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).