Equations of Straight Lines (y = mx + c) (Cambridge (CIE) IGCSE International Maths)

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Finding Equations of Straight Lines

What is the equation of a straight line?

  • The general equation of a straight line is y  = mx  + c  where

    • m  is the gradient

    • c  is the y-intercept

      • The value where it cuts the y-axis

  • y  = 5x  + 2  is a straight line with

    • gradient 5

    • y-intercept 2

  • y  = 3 - 4x  is a straight line with

    • gradient -4

    • y-intercept 3

How do I find the equation of a straight line from a graph?

  • Find the gradient by drawing a triangle and using

    • gradient equals rise over run

      • Positive for uphill lines, negative for downhill

  • Read off the y-intercept from the graph

    • Where it cuts the y-axis

  • Substitute these values into y  = mx  + c 

What if no y-intercept is shown?

  • If you can't read off the y-intercept

    • find any point on the line

    • substitute it into the equation

    • solve to find c 

  • For example, a line with gradient 6 passes through (2, 15)

    • The y-intercept is unknown

      • Write y  = 6x  + c

    • Substitute in x  = 2 and y  = 15

      • 15 = 6 × 2 + c

      • 15 = 12 + c

    • Solve for c

      • = 3

    • The equation is = 6x  + 3

What are the equations of horizontal and vertical lines?

  • A horizontal line has the equation y  = c

    • c  is the y-intercept

  • A vertical line has the equation = k

    •  k  is the x-intercept

  • For example

    • y = 4

    • x = -2

Worked Example

(a) Find the equation of the straight line shown in the diagram below.

Graph of a straight line with negative gradient

Find m, the gradient
Identify any two points the line passes through and work out the rise and run

Line passes through (2, 4) and (10, 0)

Finding the equation of a straight line from a graph

The rise is 4
The run is 8

Calculate the fraction rise over run

rise over run equals 4 over 8 equals 1 half

The slope is downward (downhill), so it is a negative gradient

gradient, m equals negative 1 half

Now find the y-intercept
The line cuts the y-axis at 5

y-intercept,  table attributes columnalign right center left columnspacing 0px end attributes row c equals 5 end table

Substitute the gradient, m, and the y-intercept, c, into = mx  + c

y equals negative 1 half x plus 5

(b) Find the equation of the straight line with a gradient of 3 that passes through the point (5, 4).

Substitute = 3 into y  = mx  + c
Leave c  as an unknown letter

y equals 3 x plus c

Substitute = 5 and = 4 into the equation
Solve the equation to find c

table attributes columnalign right center left columnspacing 0px end attributes row 4 equals cell 3 cross times 5 plus c end cell row 4 equals cell 15 plus c end cell row cell negative 11 end cell equals c end table


You now know c
Replace c  with −11 to complete the equation of the line

y  = 3 − 11

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Mark Curtis

Author: Mark Curtis

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Mark graduated twice from the University of Oxford: once in 2009 with a First in Mathematics, then again in 2013 with a PhD (DPhil) in Mathematics. He has had nine successful years as a secondary school teacher, specialising in A-Level Further Maths and running extension classes for Oxbridge Maths applicants. Alongside his teaching, he has written five internal textbooks, introduced new spiralling school curriculums and trained other Maths teachers through outreach programmes.

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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.