IGCSEMathsEdexcelMaths A (Modular)Higher Unit 1Revision NotesGraphs & CoordinatesLinear Graphs y = mx + cPerpendicular Lines

Perpendicular Lines (Edexcel IGCSE Maths A (Modular)): Revision Note

Exam code: 4XMAF/4XMAH

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

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

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

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    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 Subject 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.