Equations of a Straight Line (Edexcel IGCSE Further Pure Maths): Revision Note

Exam code: 4PM1

Dan Finlay

Written by: Dan Finlay

Reviewed by: Lucy Kirkham

Updated on

Equations of a Straight Line

How do I find the gradient of a straight line?

  • Find two points that the line passes through with coordinates (x1, y1) and (x2, y2)

  • The gradient m  between these two points is calculated by

m=y2y1x2x1 

  • This is sometimes known as rise over run

  • The gradient of a straight line measures its slope

    • A line with gradient 1 will go up 1 unit for every unit it goes to the right

    • A line with gradient -2 will go down two units for every unit it goes to the right

What are the equations of a straight line?

  •  y=mx+c

    • This is sometimes called gradient-intercept form

    • It clearly shows the gradient m and the y-intercept (0, c)

  •  yy1=m(xx1)

    • This is sometimes called the point-gradient form

    • It clearly shows the gradient m and a point on the line (x1, y1)

  •  ax+by=c

    • This is sometimes called the general form

    • You can quickly get the x-intercept (ca, 0)y-intercept (0, cb)  and gradient m=ab

      • c is not the y-intercept in this form of the line equation!

    • You can also rearrange the equation into gradient-intercept form

      • y=abx+cb

How do I find an equation of a straight line?

  • You will need the gradient

    • If you are given two points then first find the gradient

  • It is easiest to start with the point-gradient form  yy1=m(xx1)

    • then rearrange into whatever form is required

      • multiplying both sides by any denominators will get rid of fractions

  • Always check your answer

    • Substitute the coordinates of points on the line into the equation 

    • Make sure the equation is satisfied with those coordinates

Examiner Tips and Tricks

  • Make sure you state equations of straight lines in the form required

    • Usually  y=mx+c  or  ax+by=c

    • Check whether coefficients need to be integers

      • This is often the case for  ax+by=c

Worked Example

The line  l passes through the points (2, 5) and (6, 7).

Find the equation of  l , giving your answer in the form  ax+by=c where  a, b and c are integers to be found.


First find the gradient of the line using 'rise over run'

m=756(2)=128=32

Substitute the gradient and the coordinates of one of the points into   yy1=m(xx1)

y5=32(x(2))y5=32(x+2)y5=32x3

Multiply both sides of the equation by 2 to get rid of the fraction

2(y5)=2(32x3)2y10=3x6

Rearrange into the required form

3x+2y=4

Parallel Lines

How are the equations of parallel lines connected?

  • Parallel lines are always equidistant meaning they never intersect

  • Parallel lines have the same gradient

    • If the gradient of line l1 is m1 and the gradient of line l2 is m2 then

      • If m1=m2 then l1 and l2 are parallel

      • If l1 and l2 are parallel, then m1=m2

  • To determine if two lines are parallel:

    • Rearrange into the form  y=mx+c

    • Compare the gradients (i.e. the coefficients of  x)

    • If they are equal then the lines are parallel

Parallel & Perpendicular Gradients Notes Diagram 1

Worked Example

The line  l  passes through the point  (4,1)  and is parallel to the line with equation  2x5y=3 .

Find the equation of  l , giving your answer in the form  y=mx+c.


First find the gradient of the given line from its equation

2x5y=35y=2x3y=25x35

So the gradient of line l is 25

Insert that gradient and the coordinates of the point into  yy1=m(xx1)

y(1)=25(x4)y+1=25x85

Rearrange into the form required

y=25x135

Perpendicular Lines

How are the equations of perpendicular lines connected?

  • Perpendicular lines intersect at right angles

  • The gradients of two perpendicular lines are negative reciprocals

    • This means the product of their gradients is equal to 1

      • e.g. 12 and 2

    • If the gradient of line l1 is m1 and the gradient of line l2 is m2 then...

      • If m1×m2=1 then l1 and l2 are perpendicular

      • If l1 and l2 are perpendicular, then m1×m2=1

  • To determine if two lines are perpendicular:

    • Rearrange into the form  y=mx+c

    • Compare the gradients (i.e. the coefficients of x)

    • If their product is 1 then the lines are perpendicular

  • Be careful with horizontal and vertical lines

    •  x=p and  y=q are perpendicular where p and q are constants

Parallel & Perpendicular Gradients Notes Diagram 3

Worked Example

The line l1 is given by the equation  3x5y=7.

The line l2 is given by the equation  y=1453x .

Determine whether  l1 and  l2 are perpendicular. Give a reason for your answer.

Rearrange the l1 equation into y=mx+c form

3x5y=75y=3x7y=35x75

So the gradient of l1 is 35

The gradient of l2 is 53  (don't be fooled by the order of the terms in the equation!)
Find the product of the two gradients

35×(53)=1515=1

The product of the gradients is equal to 1, so the lines are perpendicular

The product of the gradients of the two lines is equal to 1, so l1 and l2 are perpendicular

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Dan Finlay

Author: Dan Finlay

Expertise: Portfolio Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

Lucy Kirkham

Reviewer: Lucy Kirkham

Expertise: Content Creator

Lucy has been a passionate Maths teacher for over 12 years, teaching maths across the UK and abroad helping to engage, interest and develop confidence in the subject at all levels.Working as a Head of Department and then Director of Maths, Lucy has advised schools and academy trusts in both Scotland and the East Midlands, where her role was to support and coach teachers to improve Maths teaching for all.