Coordinate Geometry | OCR GCSE Maths Revision Notes 2022

What is coordinate geometry?

Coordinate geometry is the study of geometric figures like lines and shapes, using coordinates.
Given two points, at GCSE, you are expected to know how to find Gradient of a Line using the formula below.
It is also useful to know how to find the Midpoint and Length of a Line Segment, and the methods for these are also shown below.

Gradient of a Line

What is the gradient of a line?

The gradient is a measure of how steep a 2D line is

A large value for the gradient means the line is steeper than for a small value of the gradient

A gradient of 3 is steeper than a gradient of 2
A gradient of −5 is steeper than a gradient of −4

A positive gradient means the line goes upwards from left to right
A negative gradient means the line goes downwards from left to right

In the equation for a straight line, $y = m x + c$ , the gradient is represented by $m$

The gradient of $y = - 3 x + 2$ is −3

How do I find the gradient of a line?

The gradient can be calculated using

$gradient = \frac{change in y}{change in x}$

You may see this written as $\frac{rise}{run}$ instead
For two coordinates $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ the gradient of the line joining them is

$\frac{y_{2} - y_{1}}{x_{2} - x_{1}} or \frac{y_{1} - y_{2}}{x_{1} - x_{2}}$

The order of the coordinates must be consistent on the top and bottom
i.e. (Point 1 – Point 2) or (Point 2 – Point 1) for both the top and bottom

How do I draw a line with a given gradient?

A line with a gradient of 4 could instead be written as $\frac{4}{1}$ .
- As $gradient = \frac{change in y}{change in x}$ , this would mean for every 1 unit to the right ( $x$ direction), the line moves upwards ( $y$ direction) by 4 units.
- Notice that 4 also equals $\frac{- 4}{- 1}$ , so for every 1 unit to the left, the line moves downwards by 4 units
If the gradient was −4, then $\frac{rise}{run} = \frac{- 4}{+ 1}$ or $\frac{+ 4}{- 1}$ . This means the line would move downwards by 4 units for every 1 unit to the right.
If the gradient is a fraction, for example $\frac{2}{3}$ , we can think of this as either
- For every 1 unit to the right, the line moves upwards by $\frac{2}{3}$ , or
- For every 3 units to the right, the line moves upwards by 2.
- (Or for every 3 units to the left, the line moves downwards by 2.)
If the gradient was $- \frac{2}{3}$ this would mean the line would move downwards by 2 units for every 3 units to the right
Once you know this, you can select a point (usually given, for example the $y$ -intercept) and then count across and upwards or downwards to find another point on the line, and then join them with a straight line

Exam Tip

Be very careful with negative numbers when calculating the gradient; write down your working rather than trying to do it in your head to avoid mistakes
- For example, $\frac{(- 3) - (- 9)}{(- 18) - (7)}$

Worked example

(a)

Find the gradient of the line joining (-1, 4) and (7, 28)

Using $gradient = \frac{change in y}{change in x}$ :

$\frac{28 - 4}{7 - - 1}$

Simplify:

$\frac{28 - 4}{7 - - 1} = \frac{24}{8} = 3$

Gradient = 3

(b)

On the grid below, draw a line with gradient −2 that passes through (0, 1).

Mark the point (0, 1) and then count 2 units down for every 1 unit across

cie-igcse-gradients-of-lines-we-1

On the grid below, draw a line with gradient $\frac{2}{3}$ that passes through (0,-1)

Mark the point (0,-1) and then count 2 units up for every 3 units across

cie-igcse-gradients-of-lines-we-2

Midpoint of a Line

How do I find the midpoint of a line in two dimensions (2D)?

The midpoint of a line will be the same distance from both endpoints
You can think of a midpoint as being the average (mean) of two coordinates
The midpoint of $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Exam Tip

Making a quick sketch of the two points will help you know roughly where the midpoint should be, which can be helpful to check your answer

Worked example

The coordinates of A are (−4, 3) and the coordinates of B are (8, −12).
Find M, the midpoint of AB.

The midpoint can be found using M = $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$ - or taking the average of the two x-coordinates and the two y-coordinates

$(\frac{- 4 + 8}{2}, \frac{3 + - 12}{2}) = (\frac{4}{2}, \frac{- 9}{2})$

Simplify

M = (2, −4.5)

Length of a Line

How do I calculate the length of a line?

The distance between two points with coordinates $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found using the formula

$d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

This formula is really just Pythagoras’ Theorem $a^{2} = b^{2} + c^{2}$ , applied to the difference in the $x$ -coordinates and the difference in the $y$ -coordinates;

Basic Coordinate Geometry Notes Diagram 2

You may be asked to find the length of a diagonal in 3D space. This can be answered using 3D Pythagoras

Exam Tip

Sketch the points and add a third point to make a right angle triangle, then use Pythagoras' Theorem.

Worked example

Point A has coordinates (3, -4) and point B has coordinates (-5, 2).

Calculate the distance of the line segment AB.

Sketch the points and form a right-angle triangle

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Use Pythagoras' Theorem, $d = \sqrt{x^{2} + y^{2}}$ , to find distance between A and B

Substituting in the two given coordinates:

$x = 3 - (- 5) = 8 y = 2 - (- 4) = 6$

Simplify:

$d = \sqrt{{(8)}^{2} + 6^{2}} = \sqrt{64 + 36} = \sqrt{100} = 10$

Answer = 10 units

Coordinate Geometry (OCR GCSE Maths)

Revision Note

What is coordinate geometry?

Gradient of a Line

What is the gradient of a line?

How do I find the gradient of a line?

How do I draw a line with a given gradient?

Exam Tip

Worked example

Midpoint of a Line

How do I find the midpoint of a line in two dimensions (2D)?

Exam Tip

Worked example

Length of a Line

How do I calculate the length of a line?

Exam Tip

Worked example

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Author: Daniel I