Equations of a Straight Line — Edexcel A-level Mathematics (Pure)

SPEC 3.1 - COORDINATE GEOMETRY

Equations of Straight Lines

Master the three key forms of straight line equations

Why Straight Lines Matter

Straight lines are the foundation of coordinate geometry. Understanding their equations allows us to model real-world relationships and solve geometric problems.

Key Concept

Every straight line can be described by an equation linking the x and y coordinates of points on the line.

In This Lesson

Three forms of straight line equations

Finding equations from given information

Converting between different forms

Exam Tip

Questions often ask for your answer "in the form ax + by + c = 0" - this means integers only, no fractions!

Gradient Recap

The gradient (or slope) measures how steep a line is. It tells us how much y changes for each unit change in x.

Formula

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Rise over Run

Positive Gradient

Line slopes upward (left to right)

m > 0

Negative Gradient

Line slopes downward (left to right)

m < 0

Zero Gradient

Horizontal line

m = 0

Important

Vertical lines have undefined gradient (division by zero). Their equation is x = k.

Form 1: y = mx + c

Gradient-Intercept Form

y = m x + c

SymbolMeaningExample

mGradient of the lineIf m = 3, line rises 3 for every 1 across

cy-intercept (where line crosses y-axis)If c = -2, line crosses y-axis at (0, -2)

When to Use

Use this form when you know the gradient and the y-intercept, or when you need to quickly identify these values from an equation.

Example

y = 2 x + 5

has gradient 2 and crosses the y-axis at (0, 5)

Worked Example01

Find the equation of the straight line with gradient 3 and y-intercept -2.

1

Identify the values: m = 3, c = -2

2

Substitute into y = mx + c

3

y = 3 x + (- 2)

Answer: y = 3x - 2

Form 2: y - y₁ = m(x - x₁)

Point-Gradient Form

y - y_{1} = m (x - x_{1})

SymbolMeaning

mGradient of the line

(x₁, y₁)Any known point on the line

When to Use

Use this form when you know the gradient and any point on the line (not necessarily the y-intercept). This is the most versatile form!

Common Mistake

Be careful with signs! If your point is (3, -4), then y₁ = -4, so you get y - (-4) = y + 4

Worked Example02

Find the equation of the straight line with gradient 2 that passes through the point (3, 5).

1

Identify: m = 2, point (x₁, y₁) = (3, 5)

2

Substitute into y - y₁ = m(x - x₁)

3

y - 5 = 2 (x - 3)

4

Expand:

y - 5 = 2 x - 6

5

Rearrange:

y = 2 x - 6 + 5

Answer: y = 2x - 1

Worked Example03

Find the equation of the straight line passing through the points (1, 4) and (3, 10).

1

Find the gradient:

m = \frac{10 - 4}{3 - 1} = \frac{6}{2} = 3

2

Use point-gradient form with point (1, 4):

3

y - 4 = 3 (x - 1)

4

Expand:

y - 4 = 3 x - 3

5

Rearrange:

y = 3 x - 3 + 4

Answer: y = 3x + 1

Form 3: ax + by + c = 0

General Form

a x + b y + c = 0

Key Points

a, b, and c are integers (whole numbers)

Usually we make 'a' positive

There are no fractions in the final answer

All terms are on one side, with 0 on the other

Exam Requirement

When a question says "give your answer in the form ax + by + c = 0", you MUST use integers. Multiply through to clear any fractions!

Example

3 x - 2 y + 7 = 0

is in general form (a = 3, b = -2, c = 7)

Worked Example04

Convert $y = \frac{2}{3} x - 4$ to the form ax + by + c = 0

1

Start with:

y = \frac{2}{3} x - 4

2

Multiply everything by 3 to clear the fraction:

3 y = 2 x - 12

3

Rearrange to get 0 on one side:

2 x - 3 y - 12 = 0

4

Check: a = 2 is positive, all integers

Answer: 2x - 3y - 12 = 0

Worked Example05

Find the equation of the line through (-2, 5) and (4, -1), giving your answer in the form ax + by + c = 0.

1

Find gradient:

m = \frac{- 1 - 5}{4 - ( - 2 )} = \frac{- 6}{6} = - 1

2

Use point-gradient form with (-2, 5):

y - 5 = - 1 (x - (- 2))

3

Simplify:

y - 5 = - 1 (x + 2)

4

Expand:

y - 5 = - x - 2

5

Rearrange:

x + y - 5 + 2 = 0

Answer: x + y + 3 = 0

Summary: Three Forms

FormEquationBest Used When

Gradient-Intercept

y = m x + c

You know gradient and y-intercept

Point-Gradient

y - y_{1} = m (x - x_{1})

You know gradient and any point

General

a x + b y + c = 0

Exam requires integer form

Strategy

For most problems: use point-gradient form first, then convert to the required form at the end.

Practice

Question 1: Find the equation of the line with gradient 4 passing through the point (2, 7). Give your answer in the form y = mx + c.

1

Use y - y₁ = m(x - x₁) with m = 4, (x₁, y₁) = (2, 7)

2

y - 7 = 4 (x - 2)

3

y - 7 = 4 x - 8

4

y = 4 x - 8 + 7

Answer: y = 4x - 1

Practice

Question 2: Find the equation of the line through (-3, 2) and (5, 6), giving your answer in the form ax + by + c = 0.

1

Gradient:

m = \frac{6 - 2}{5 - ( - 3 )} = \frac{4}{8} = \frac{1}{2}

2

Using (-3, 2):

y - 2 = \frac{1}{2} (x + 3)

3

Multiply by 2:

2 (y - 2) = x + 3

4

2 y - 4 = x + 3

5

Rearrange:

x - 2 y + 7 = 0

Answer: x - 2y + 7 = 0

Practice

Question 3: The line L has equation 5x - 2y + 8 = 0. Find the gradient of L and the coordinates of the point where L crosses the y-axis.

1

Rearrange:

- 2 y = - 5 x - 8

2

Divide by -2:

y = \frac{5}{2} x + 4

3

Compare with y = mx + c

4

Gradient: m = 5/2

5

y-intercept: c = 4, so crosses y-axis at (0, 4)

Gradient = 5/2 (or 2.5)

Crosses y-axis at (0, 4)

Summary

Key Formulas

Gradient

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Point-Gradient

y - y_{1} = m (x - x_{1})

Gradient from general form

m = - \frac{a}{b}

for ax + by + c = 0

Exam Checklist

Can find gradient from two points

Can write equation given gradient and point

Can write equation through two points

Can convert between all three forms

Can find gradient and intercept from any form