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

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 - 4*x 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

  • $\mathrm{gradient}=\frac{\mathrm{rise}}{\mathrm{run}}$

    • 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

    • c = 3

  • The equation is y = 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 x = 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.

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)

The rise is 4 The run is 8

Calculate the fraction $\frac{\mathrm{rise}}{\mathrm{run}}$

$\frac{\mathrm{rise}}{\mathrm{run}}=\frac{4}{8}=\frac{1}{2}$

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

gradient, $m=-\frac{1}{2}$

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

y-intercept, *$\begin{array}{rcl} c = 5 \end{array}$

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

$y=-\frac{1}{2}x+5$

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

Answer:

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

$y=3x+c$

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

$\begin{array}{rcl} 4 = 3 \times 5 + c \\ 4 = 15 + c \\ - 11 & = c \end{array}$

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

$y=3x-11$