Equations of Straight Lines (y = mx + c) (WJEC GCSE Maths & Numeracy (Double Award): Foundation): Revision Note

Exam code: 3320

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=riserun

      • 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

Answer:

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 riserun

riserun=48=12

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

gradient, m=12

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

y-intercept,  c=5

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

y=12x+5

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

Answer:

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

y=3x+c

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

4=3×5+c4=15+c11=c


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

y=3x11

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