Basic Coordinate Geometry

What are Cartesian coordinates?

Cartesian coordinates are the coordinates used in the x-y coordinate system
- They allow us to label where things are in a two-dimensional plane
In the 2D Cartesian system, the horizontal axis is labelled x and the vertical axis is labelled y

What can I do with coordinates?

If you have two points with coordinates $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ then you should be able to find
- The distance between the two points
- The point that divides the line between the two points in a given ratio
- The gradient of the line joining the two points

How do I find the distance between two points?

The distance $d$ between two points with coordinates $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found using the formula
- $d = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

$d$ is the hypotenuse of a right-angled triangle
- The differences between the $x$ -coordinates and the $y$ -coordinates give the other two side lengths
- So by Pythagoras’ Theorem $d^{2} = {(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}$

How do I divide the line between two points in a given ratio?

The coordinates of the point dividing the line joining $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ in the ratio $m : n$ are given by
- $(\frac{n x_{1} + m x_{2}}{m + n}, \frac{n y_{1} + m y_{2}}{m + n})$

- - If $m > n$ then the point is closer to $(x_{2}, y_{2})$
  - If $m < n$ then the point is closer to $(x_{1}, y_{1})$

How do I find the midpoint between two points?

The midpoint is the point halfway between the two points
- This is just 'dividing in a given ratio' where the ratio is $1 : 1$
The coordinates of the midpoint will be
- $(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Midpoint of a line segment

How do I find the gradient of a straight line joining two points?

The gradient $m$ of a line between two points with coordinates $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ can be found using the formula
- $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

- - This is also known as $m = \frac{rise}{run}$

Worked example

Point $A$ has coordinates $(3, - 4)$ and point $B$ has coordinates $(- 5, 8)$ .

(a)

Calculate the distance between points $A$ and $B$ .

Let

A (3, - 4)

be 'point 1', so

x_{1} = 3

and

y_{1} = - 4

Let

B (- 5, 8)

be 'point 2', so

x_{2} = - 5

and

y_{2} = 8

Use the formula

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

d = \sqrt{{(3 - (- 5))}^{2} + {(- 4 - 8)}^{2}} = \sqrt{8^{2} + {(- 12)}^{2}} = \sqrt{64 + 144}

\sqrt{208}

(or

4 \sqrt{13}

or 14.4 to 3 s.f.)

(b)

Find the gradient of the line joining points A and B.

Let

A (3, - 4)

be 'point 1', so

x_{1} = 3

and

y_{1} = - 4

Let

B (- 5, 8)

be 'point 2', so

x_{2} = - 5

and

y_{2} = 8

Use the formula

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

m = \frac{8 - (- 4)}{- 5 - 3} = \frac{12}{- 8}

Simplify the fraction

- \frac{3}{2}

(c)

Find point that divides the line joining points A and B in the ratio

1 : 3

Let

A (3, - 4)

be 'point 1', so

x_{1} = 3

and

y_{1} = - 4

Let

B (- 5, 8)

be 'point 2', so

x_{2} = - 5

and

y_{2} = 8

Use the formula

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

with

m = 1

and

n = 3

(\frac{(3 (3) + 1 (- 5))}{1 + 3}, \frac{(3 (- 4) + 1 (8))}{1 + 3}) = (\frac{9 - 5}{4}, \frac{- 12 + 8}{4})

(1, - 1)

Basic Coordinate Geometry (Edexcel IGCSE Further Maths)

Revision Note