IGCSEMathsCambridge (CIE)ExtendedRevision NotesCoordinate Geometry & GraphsLinear GraphsPerpendicular Lines

Perpendicular Lines (Cambridge (CIE) IGCSE Maths) : Revision Note

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

Cambridge (CIE) Maths: ExtendedEdexcel Maths A (Modular): Higher Unit 1Edexcel Maths A: HigherCambridge (CIE) International Maths: Extended

Did this video help you?

Perpendicular Lines

What are perpendicular lines?

  • Perpendicular lines are straight lines which meet at right-angles (90°)

  • One line may be referred to as a normal to the other line

  • Gradients m1 and m2 are perpendicular if m1 × m2 = −1

    • For example

      • 1 and −1

      • 1 third and -3

      • negative 2 over 3 and 3 over 2

  • The two gradients are negative reciprocals of one another

  • We can use m subscript 2 equals negative 1 over m subscript 1to find a perpendicular gradient

How can I tell if two lines are perpendicular?

  • Given two lines in the form y equals m x plus c, simply check if their gradients open parentheses m close parentheses are negative reciprocals of one another

    • y equals 1 third x plus 10 and y equals negative 3 x minus 18 are perpendicular

    • y equals 1 over 7 x plus 16 and y equals 7 x minus 8 are not perpendicular

  • One or both of the equations may not be written in the form y equals m x plus c

    • In this case, you should rearrange both equations into the form bold italic y bold equals bold italic m bold italic x bold plus bold italic c

    • Their gradients can then be easily compared

How do I find the equation of a line perpendicular to another?

  • You need to be able to find the equation of line that passes through a particular point and is perpendicular to another line

    • E.g. 5 y equals 4 x plus 30 which passes through the point (8, 3)

  • Rearrange the equation into the form y equals m x plus c so that its gradient can be identified more easily

    • y equals 4 over 5 x plus 6

  • Find the gradient of the perpendicular line

    • The gradient of the original line is 4 over 5

    • Therefore the gradient of the perpendicular line is negative 5 over 4

    • The perpendicular line has an equation in the form y equals negative 5 over 4 x plus c

  • Substitute the given point into the equation for the perpendicular and solve for c

    • Substitute (8, 3), into y equals negative 5 over 4 x plus c

    • 3 equals negative 5 over 4 open parentheses 8 close parentheses plus c

    • c equals 13

  • Substitute the value of c to find the equation of the perpendicular

    • The equation of the perpendicular line is y equals negative 5 over 4 x plus 13

      • This could also be written as 4 y equals negative 5 x plus 52 or equivalent

Worked Example

The line L has equation y minus 2 x plus 2 equals 0. 

Find the equation of the line perpendicular to L which passes through the point open parentheses 2 comma space minus 3 close parentheses.

Leave your answer in the form a x plus b y plus c equals 0 where a, b and c are integers.

Rearrange L into the form y equals m x plus c so we can identify the gradient

table row cell y minus 2 x plus 2 end cell equals 0 row y equals cell 2 x minus 2 end cell end table

Gradient of L is 2

The gradient of the line perpendicular to L will be the negative reciprocal of 2

m equals negative 1 half

Substitute the point open parentheses 2 comma space minus 3 close parentheses into the equation y equals negative 1 half x plus c
Solve for c

table row y equals cell negative 1 half x plus c end cell row cell negative 3 end cell equals cell negative 1 half open parentheses 2 close parentheses plus c end cell row cell negative 3 end cell equals cell negative 1 plus c end cell row c equals cell negative 2 end cell end table

Write the full equation of the line

table row y equals cell negative 1 half x minus 2 end cell end table

The question asks for the line to be written in the form a x plus b y plus c equals 0 where a, b and c are integers

Move all the terms to the left hand side

1 half x plus y plus 2 equals 0

Then multiply every term by 2, to ensure they are all integers

bold italic x bold plus bold 2 bold italic y bold plus bold 4 bold equals bold 0

How do I find the equation of a perpendicular bisector?

  • A perpendicular bisector of a line segment cuts the line segment in half at a right angle

  • Finding the equation of the perpendicular bisector of a line segment is very similar to finding the equation of a any perpendicular

    • Find the coordinates of the midpoint of the line segment

      • The perpendicular bisector will pass through this point

    • Find the gradient of the line segment

    • Then find the negative reciprocal of this gradient

      • This will be the gradient of the perpendicular bisector, m

    • Write the equation of the perpendicular bisector in the form y equals m x plus c

    • Substitute the midpoint of the line segment into the equation of the perpendicular bisector

      • Solve to find c

    • Write the full equation of the perpendicular bisector in the form y equals m x plus c

    • Rearrange the equation if the question requires a different form

Worked Example

Find the equation of the perpendicular bisector of the line segment joining the points (4, -6) and (8, 6).

Find the coordinates of the midpoint of the line segment
The perpendicular bisector will pass through this point

open parentheses fraction numerator 4 plus 8 over denominator 2 end fraction comma space fraction numerator negative 6 plus 6 space over denominator 2 end fraction close parentheses space equals space open parentheses 6 comma space 0 close parentheses

Find the gradient of the line segment

fraction numerator negative 6 minus 6 over denominator 4 minus 8 end fraction equals fraction numerator negative 12 over denominator negative 4 end fraction equals 3

Find the negative reciprocal of this
This will be the gradient of the perpendicular bisector, m

m equals negative 1 third

Write the equation of the perpendicular bisector in the form y equals m x plus c

y equals negative 1 third x plus c

Substitute in the midpoint (6, 0) and solve to find c

table row 0 equals cell negative 1 third open parentheses 6 close parentheses plus c end cell row 0 equals cell negative 2 plus c end cell row c equals 2 end table

Write the full equation of the perpendicular bisector

bold italic y bold equals bold minus bold 1 over bold 3 bold italic x bold plus bold 2

👀 You've read 1 of your 5 free revision notes this week
An illustration of students holding their exam resultsUnlock more revision notes. It's free!

By signing up you agree to our Terms and Privacy Policy.

Already have an account? Log in

Test yourselfFlashcards

Did this page help you?

Previous:Parallel LinesNext:Quadratic Graphs
Author
Reviewer
Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths 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.

Download notes on Perpendicular Lines