Equations of a Straight Line (Edexcel IGCSE Further Maths)

Revision Note

Author

Dan

Expertise

Maths

Equations of a Straight Line

How do I find the gradient of a straight line?

Find two points that the line passes through with coordinates (x₁, y₁) and (x₂, y₂)
The gradient m between these two points is calculated by

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

- This is sometimes known as rise over run
The gradient of a straight line measures its slope
- A line with gradient 1 will go up 1 unit for every unit it goes to the right
- A line with gradient -2 will go down two units for every unit it goes to the right

What are the equations of a straight line?

$y = m x + c$
- This is sometimes called gradient-intercept form
- It clearly shows the gradient m and the y-intercept (0, c)
$y - y_{1} = m (x - x_{1})$
- This is sometimes called the point-gradient form
- It clearly shows the gradient m and a point on the line (x₁, y₁)
$a x + b y = c$
- This is sometimes called the general form
- You can quickly get the x-intercept $(\frac{c}{a}, 0)$ , y-intercept $(0, \frac{c}{b})$ and gradient $m = - \frac{a}{b}$
  - $c$ is not the y-intercept in this form of the line equation!
- You can also rearrange the equation into gradient-intercept form
  - $y = - \frac{a}{b} x + \frac{c}{b}$

How do I find an equation of a straight line?

You will need the gradient
- If you are given two points then first find the gradient
It is easiest to start with the point-gradient form $y - y_{1} = m (x - x_{1})$
- then rearrange into whatever form is required
  - multiplying both sides by any denominators will get rid of fractions
Always check your answer
- Substitute the coordinates of points on the line into the equation
- Make sure the equation is satisfied with those coordinates

Exam Tip

Make sure you state equations of straight lines in the form required
- Usually $y = m x + c$ or $a x + b y = c$
- Check whether coefficients need to be integers
  - This is often the case for $a x + b y = c$

Worked example

The line $l$ passes through the points $(- 2, 5)$ and $(6, - 7)$ .

Find the equation of $l$ , giving your answer in the form $a x + b y = c$ where $a, b$ and $c$ are integers to be found.

First find the gradient of the line using 'rise over run'

$m = \frac{- 7 - 5}{6 - (- 2)} = \frac{- 12}{8} = - \frac{3}{2}$

Substitute the gradient and the coordinates of one of the points into $y - y_{1} = m (x - x_{1})$

$\begin{array}{rcl} y - 5 & = & - \frac{3}{2} (x - (- 2)) \\ y - 5 & = & - \frac{3}{2} (x + 2) \\ y - 5 & = & - \frac{3}{2} x - 3 \end{array}$

Multiply both sides of the equation by 2 to get rid of the fraction

$\begin{array}{rcl} 2 (y - 5) & = & 2 (- \frac{3}{2} x - 3) \\ 2 y - 10 & = & - 3 x - 6 \end{array}$

Rearrange into the required form

$3 x + 2 y = 4$

Parallel Lines

How are the equations of parallel lines connected?

Parallel lines are always equidistant meaning they never intersect
Parallel lines have the same gradient
- If the gradient of line is and the gradient of line is then
  - If $m_{1} = m_{2}$ then $l_{1}$ and $l_{2}$ are parallel
  - If $l_{1}$ and $l_{2}$ are parallel, then $m_{1} = m_{2}$
To determine if two lines are parallel:
- Rearrange into the form $y = m x + c$
- Compare the gradients (i.e. the coefficients of $x$ )
- If they are equal then the lines are parallel

Parallel & Perpendicular Gradients Notes Diagram 1

Worked example

The line $l$ passes through the point $(4, - 1)$ and is parallel to the line with equation $2 x - 5 y = 3$ .

Find the equation of $l$ , giving your answer in the form $y = m x + c$ .

First find the gradient of the given line from its equation

$\begin{array}{rcl} 2 x - 5 y & = & 3 \\ 5 y & = & 2 x - 3 \\ y & = & \frac{2}{5} x - \frac{3}{5} \end{array}$

So the gradient of line $l$ is $\frac{2}{5}$

Insert that gradient and the coordinates of the point into $y - y_{1} = m (x - x_{1})$

$\begin{array}{rcl} y - (- 1) & = & \frac{2}{5} (x - 4) \\ y + 1 & = & \frac{2}{5} x - \frac{8}{5} \end{array}$

Rearrange into the form required

$y = \frac{2}{5} x - \frac{13}{5}$

Perpendicular Lines

How are the equations of perpendicular lines connected?

Perpendicular lines intersect at right angles
The gradients of two perpendicular lines are negative reciprocals
- This means the product of their gradients is equal to
  - e.g. $\frac{1}{2}$ and $- 2$
- If the gradient of line is and the gradient of line is then...
  - If $m_{1} \times m_{2} = - 1$ then $l_{1}$ and $l_{2}$ are perpendicular
  - If $l_{1}$ and $l_{2}$ are perpendicular, then $m_{1} \times m_{2} = - 1$
To determine if two lines are perpendicular:
- Rearrange into the form $y = m x + c$
- Compare the gradients (i.e. the coefficients of $x$ )
- If their product is $- 1$ then the lines are perpendicular
Be careful with horizontal and vertical lines
- $x = p$ and $y = q$ are perpendicular where p and q are constants

Parallel & Perpendicular Gradients Notes Diagram 3

Worked example

The line $l_{1}$ is given by the equation $3 x - 5 y = 7$ .

The line $l_{2}$ is given by the equation $y = \frac{1}{4} - \frac{5}{3} x$ .

Determine whether $l_{1}$ and $l_{2}$ are perpendicular. Give a reason for your answer.

Rearrange the $l_{1}$ equation into $y = m x + c$ form

$\begin{array}{rcl} 3 x - 5 y & = & 7 \\ 5 y & = & 3 x - 7 \\ y & = & \frac{3}{5} x - \frac{7}{5} \end{array}$

So the gradient of $l_{1}$ is $\frac{3}{5}$

The gradient of $l_{2}$ is $- \frac{5}{3}$ (don't be fooled by the order of the terms in the equation!)
Find the product of the two gradients

$\frac{3}{5} \times (- \frac{5}{3}) = - \frac{15}{15} = - 1$

The product of the gradients is equal to $- 1$ , so the lines are perpendicular

The product of the gradients of the two lines is equal to $- 1$ , so $l_{1}$ and $l_{2}$ are perpendicular

You've read 0 of your 0 free revision notes

Get unlimited access

to absolutely everything:

Downloadable PDFs
Unlimited Revision Notes
Topic Questions
Past Papers
Model Answers
Videos (Maths and Science)

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Next topic

Did this page help you?

Equations of a Straight Line (Edexcel IGCSE Further Maths)

Revision Note

How do I find the gradient of a straight line?

What are the equations of a straight line?

How do I find an equation of a straight line?

How are the equations of parallel lines connected?

How are the equations of perpendicular lines connected?

You've read 0 of your 0 free revision notes

Get unlimited access

Join the 100,000+ Students that ❤️ Save My Exams

Author: Dan