CO-ORDINATEGEOMETRY

Mid-point of two points:

To find the mid-point, we need the middle of the x co-ordinates

i.e. the average of 1 and 7, which is

In general:

Let A be the point (1, 2)

Let B be the point (7, 4)

4

and we need the middle of the y co-ordinates

i.e. the average of 2 and 4, which is 3

y

xA

B

(1, 2)

(7, 4)

Let M be the mid-point.

M

So, M is (4, 3)

x1 + x2

2y1 + y2

2( , )The mid-point of (x1, y1) and (x2, y2) is:

Example 1: Find the mid-point of A and B, given A is the point ( –1, 4) and B is the point ( 7, – 6).

y

xx1 + x2

2y1 + y2

2( , )

The mid-point is given by:

–1+ 7 2

4 + – 6 2

( , )i.e.

Hence the mid-point is: ( 3, –1 )

A ( –1, 4)

B ( 7, – 6)

( 3, –1 )

Distance between two points:

To find the distance we use Pythagoras’ theorem.

The difference between the x co-ordinates is

Let A be the point (1, 2)

Let B be the point (13, 7)

y

x

A(1, 2)

B (13, 7)

13 – 1 = 12

Hence, the distance AB is given by:

12

5

The difference between the y co-ordinates is 7 – 2 = 5

AB2 = 122 + 52.i.e. AB = 22 512

In general:

The distance between (x1, y1) and (x2, y2) is: 212

212 )()( yyxx

= 13

Example 1: Find the distance between A and B, given A is the point ( –2, 5) and B is the point ( 6, – 10).

x

yA ( –2, 5)

B ( 6, – 10) 8

15

212

212 )()( yyxx

The distance AB is given by:

22 )510()26( =

22 )15()8( = 289= = 17

Gradient of the line joining two points:

To find the gradient we use the definition:

Let A be the point (1, 2)

Let B be the point (13, 6)

12

4

A(1,2)

B (13, 6)y

x

Gradient, m = difference in ydifference in x

= 6 – 2 13 – 1

= 412

= 13

The gradient of the line through (x1, y1) and (x2, y2) is: y2 – y1

x2 – x1

Perpendicular lines:

Let A be the point (5, 3)

If the line OA is rotated 90º anticlockwise about O, to givethe line OB,

A

y

x

(5, 3)

Othen B is the point

Now, the gradient of OA is: 3 – 0 5 – 0

= 35

and the gradient of OB is: 5 – 0 –3 – 0

=– 5 3

1m

–

If a line has gradient m, then a line perpendicular to this line has

gradient

Note that theproduct ofthese gradientsis – 1.

(–3, 5 ) B

(–3, 5).

3

3

5

5

Example 1: Find the gradient of the line through A and B, given A is the point ( 1, 5) and B is the point ( 4, – 1). Find also thegradient of the line perpendicular to the line through A and B.

y

x

A ( 1, 5)

B ( 4, – 1)

The gradient of AB is given by:

5 – – 1 1 – 4

=

= 6–3

= – 2 3

6

Note: The gradient can be found by considering the lengths foundin the sketch, and noting that the line has a negative gradient.

Gradient, m = y2 – y1

x2 – x1

The perpendicular line has gradient 1m

– 1–2

–= =12

Summary of key points:

This PowerPoint produced by R.Collins ; Updated Apr. 2009

Mid-point of two points:

The mid-point of (x1, y1) and (x2, y2) is:x1 + x2

2y1 + y2

2( , )

Distance between two points:

The distance between (x1, y1) and (x2, y2) is: 212

212 )()( yyxx

Gradient of the line joining two points:

The gradient of the line through (x1, y1) and (x2, y2) is:y2 – y1

x2 – x1

Perpendicular lines:

If a line has gradient m, then a line perpendicular to this line has

gradient 1m

–