Coordinate Geometry

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;

1. Midpoint of a Line
2. Gradient of a Line
3. Length of a Line

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

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

How can I extend the idea of the midpoint in two dimensions (2D)?

The midpoint of AB splits AB in the ratio 1 : 1.
If you are asked to find the point that divides AB in the ratio m : n, then you need to find the point that lies $\frac{m}{m + n}$ of the way from A to B.
- E.g. dividing AB in the ratio 2 : 3 means finding the point that is $\frac{2}{5}$ of the way from A to B.
Normally an exam question will ask you to find the point that divides AB in the ratio 1 : n, so;
- find the difference between the x coordinates,
- divide this difference by [1 + n], and add the result to the x coordinate of A,
- repeat for the y coordinates.

How do I find the midpoint of a line in three dimensions (3D)?

Finding the midpoint of a line in 3D involves a simple extension of the formula used for 2D midpoints
Similar to before, you can think of a midpoint as being the average (mean) of three $x$ and three $y$ coordinates
The midpoint of $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is

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

Exam Tip

If working in 2D (most questions!) 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
If working in 3D (some questions!), just check that your midpoint's $x$ coordinate lies between the two given x coordinates, and so on for the $y$ and $z$ coordinates

Worked example

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

(a)

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})

Fill in the values of x and y from each coordinate

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

Simplify

M = (2, −4.5)

(b)

A point N divides AB in the ratio 1 : 2.

Find the coordinates of N.

Calculate the difference between the x coordinate of A and the x coordinate of B

\begin{array}{rcl} 8 - (- 4) & = & 12 \end{array}

Divide this difference by 3 (as there are 1 + 2 = 3 parts in the ratio 1 : 2)

\begin{array}{rcl} 12 \div 3 & = & 4 \end{array}

Add 4 to the x coordinate of A

\begin{array}{rcl} x coordinate of N & = & - 4 + 4 = 0 \end{array}

Repeat for y

\begin{array}{rcl} - 12 - 3 & = & - 15 \\ (- 15) \div 3 & = & - 5 \\ y coordinate of N & = & 3 + (- 5) = - 2 \end{array}

Alternatively, you may find it helpful to sketch the coordinates A and B and find N intuitively. Note that in the sketch, the coordinates do not need to be placed in the correct orientation to one another, simply along a straight line:

Write the final answer as a coordinate point

N = (0, −2)

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

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

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;

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

Exam Tip

As we are squaring the difference in and in the formula, it does not matter if they are positive or negative
- 3² is the same as (-3)², this may help to speed up your working