## What you need to know

Coordinates, denoted by $(x, y)$, are what we use to communicate where a particular point is located on a pair of coordinate axes. See some examples below.

The first number in the bracket is the $x$ coordinate, and it tells us how far to move across the $x$-axis – right if it’s positive and left if it’s negative – from the origin, which is the point $(0,0)$.

The second number in the bracket is the $y$ coordinate, and it tells us how far to move up/down the $y$-axis – up if it’s positive and down if it’s negative – from the origin.

It’s a very useful and fairly straightforward system which many of you will be quite familiar with at this point. The main focus of this topic is as follows.

Imagine you have two coordinates and you draw a line between them. What coordinates do you end up at if you travel halfway along that line? What about if you go 80% of the way along that line?

To answer these questions, let’s look at some examples.

Example: Find the midpoint of the line segment which joins points $A:(2, 3)$ and $B:(10, -7)$.

The word midpoint refers to the point which is exactly halfway between the two points in question. We can also think of this as the place you’d end up if you travelled halfway along the line from one coordinate to the other. We’ll look at two approaches for finding the midpoint. In either case, we’re going to look separately at the $x$ coordinates and $y$ coordinates.

Method 1: The 2 $x$ coordinates are 2 and 10. One way to determine the half way point is to add them up and divide by 2 (in other words, take the mean of the two points).

$\text{Midpoint of }x\text{ coordinates } = (2 + 10) \div 2 = 6$

The 2 $y$ coordinates are 3 and -7. We’ll do exactly the same thing for these.

$\text{Midpoint of }y\text{ coordinates} = (-7 + 3) \div 2 = -2$

Therefore, the coordinates of the midpoint are $(6, -2)$.

Method 2: The 2 $x$ coordinates are 2 and 10. The second method for finding the midpoint is to subtract the $x$ (and then $y$) coordinate of $A$ from that of $B$, halve the answer, and add that number onto the $x$ (and then $y$) coordinate of $A$. Practically, this is like starting at $A$ and travelling halfway along the line toward $B$.

So, subtracting the $x$ coordinates gives $10 - 2 = 8$, half of which is 4. Adding this to the 2, we get that $2 + 4 = 6$ is the $x$ coordinate of the midpoint.

The $y$ coordinates are 3 and -7. Subtracting them, we get $-7 - 3 = -10$, half of which is 5. Adding this to the $y$ coordinate of $A$, we get that $-7 + 5 = -2$ is the $y$ coordinate of the midpoint.

So, as before, we find that the coordinates of the midpoint are $(6, -2)$.

The first method is more straightforward when finding the midpoint, but when faced with trickier problems like this next example, we will require the second method.

Example: Points $A$ and $B$ have coordinates $(3, 12)$ and $(-5, 16)$ respectively. Point $C$ lies on the line segment between points $A$ and $B$ such that $AC:CB = 5:3$. Find the coordinates of point $C$.

To solve this problem, we should do a method similar to method 2 from the previous example, only this time we won’t be travelling half way along the line.

In a ratio of 5:3 there are 8 parts in total, and the distance from $A$ to $C$ constitutes 5 of those parts. In other words, the distance from $A$ to $C$ counts for $\frac{5}{8}$ of the total distance between $A$ and $B$.

So, we’re still going to subtract the individual coordinates of $A$ from $B$ to find the distance in both $x$ and $y$  like we did in the previous example, but this time we’re going to find $\frac{5}{8}$ of the total (as opposed to a half) before adding the values onto the coordinates of $A$.

First, $x$ coordinates: $-5 - 3 = -8$, then $\frac{5}{8} \times -8 = -5$. Adding this to the $x$ coordinate of $A$, we get

$x\text{ coordinate of C } = 3 + (-5) = -2$

Second, $y$ coordinates: $12 - 16 = -4$, then $\frac{5}{8} \times -4 = -\frac{5}{2}$. Adding this to the $y$ coordinate of $A$, we get

$y\text{ coordinate of C } = 3 + \left(-\dfrac{5}{2}\right) = \dfrac{1}{2}$

Therefore, the coordinates of $C$ are $\left(-2, \frac{1}{2}\right)$.

## Example Questions

#### 1) Write down the coordinates of points $A, B,$ and $C$ seen below.

$A$ is -2 in the $x$ direction and 2 in the $y$ direction, so $A = (-2, 2)$.

$B$ is -1 in the $x$ direction and -2 in the $y$ direction, so $B = (-1, -2)$.

$C$ is 3 in the $x$ direction and 0 in the $y$ direction, so $C = (3, 0)$.

#### 2) Find the midpoint of the line segment that joins points $A$ and $B$ as seen below.

Point $A$ has coordinates $(-2, -2)$.

Point $B$ has coordinates $(0, 3)$.

By taking the average of the $x$ coordinates of $A$ and $B$, the $x$ coordinate of the midpoint is

$\frac{-2 + 0}{2} = -1$.

By taking the average of the $y$ coordinates of $A$ and $B$, the $y$ coordinate of the midpoint is

$\dfrac{-2 + 3}{2} = -\dfrac{1}{2}$.

Therefore, the coordinates of the midpoint are $\left(-1, -\dfrac{1}{2}\right)$.

#### 3) Points $A$ and $B$ have coordinates $(-10, 37)$ and $(-16, 1)$ respectively. Point $C$ lies on the line segment between points $A$ and $B$ such that $AC:CB = 2:7$. Find the coordinates of point $C$.

In a ratio of 2:7 there are 9 parts in total, and the distance from $A$ to $C$ constitutes 2 of those parts. Therefore, the distance from $A$ to $C$ counts for $\dfrac{2}{9}$ of the total distance between $A$ and $B$. So, we’re going to subtract the individual coordinates of $A$ from $B$ to find the distance in both $x$ and $y$, and then we are going to add $\dfrac{2}{9}$ of these respective distances to the coordinates of point $A$.

First, $x$ coordinates: $-16 -(-10) = -6$, then

$\dfrac{2}{9} \times (-6) = -\dfrac{12}{9} = -\dfrac{4}{3}$

Adding this to the $x$ coordinate of $A$, we get

$x\text{ coordinate of C } = -10 + \left(-\dfrac{4}{3}\right) = -\dfrac{34}{3}$

Second, $y$ coordinates: $1 - 37 = -36$, then

$\dfrac{2}{9} \times (-36) = -\dfrac{72}{9} = -8$

Adding this to the $y$ coordinate of $A$, we get

$y\text{ coordinate of C } = 37 + (-8) = 29$

Therefore, the coordinates of $C$ are $\left(-\frac{34}{3}, 29\right)$.

## Coordinates and Ratios Revision and Worksheets

Coordinates
Level 1-3
Mid points Ratios and Coordinates
Level 1-3

## Coordinates and Ratios Teaching Resources

If you are a GCSE Maths tutor in York or a Maths teacher in London, you will find our coordinates and ratio revision resources really useful for classroom teaching or for setting challenging homework. All of the coordinates and ratio questions are relevant tothe three major exam boards, AQA, OCR and Edexcel as well as being suitable for the new GCSE maths syllabus.