1. CO-ORDINATE GEOMETRY

2. 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 x A B (1, 2) (7, 4) Let M be the mid-point. M So, M is (4, 3) x 1 + x 2 2 y 1 + y 2 2 (, ) The mid-point of (x 1, y 1 ) and (x 2, y 2 ) is:

3. 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 x x 1 + x 2 2 y 1 + y 2 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 )

4. 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 is7 – 2 = 5 AB 2 = 12 2 + 5 2. i.e. AB = In general: The distance between (x 1, y 1 ) and (x 2, y 2 ) is: = 13

5. Example 1: Find the distance between A and B, given A is the point ( –2, 5) and B is the point ( 6, – 10). x y A ( –2, 5) B ( 6, – 10) 8 15 The distance AB is given by: = = == 17

6. 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 y difference in x = 6 – 2 13 – 1 = 4 12 = 1313 The gradient of the line through (x 1, y 1 ) and (x 2, y 2 ) is: y 2 – y 1 x 2 – x 1

7. Perpendicular lines: Let A be the point (5, 3) If the line OA is rotated 90º anticlockwise about O, to give the line OB, A y x (5, 3) O then B is the point Now, the gradient of OA is: 3 – 0 5 – 0 = 3535 and the gradient of OB is: 5 – 0 –3 – 0 = – 5 3 1m1m – If a line has gradient m, then a line perpendicular to this line has gradient Note that the product of these gradients is – 1. (–3, 5 ) B (–3, 5). 3 3 5 5

8. 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 the gradient 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 found in the sketch, and noting that the line has a negative gradient. Gradient, m = y 2 – y 1 x 2 – x 1 The perpendicular line has gradient 1m1m – 1 –2 –== 1212

9. Summary of key points: This PowerPoint produced by R.Collins ; Updated Apr. 2009 Mid-point of two points: The mid-point of (x 1, y 1 ) and (x 2, y 2 ) is: x 1 + x 2 2 y 1 + y 2 2 (, ) Distance between two points: The distance between (x 1, y 1 ) and (x 2, y 2 ) is: Gradient of the line joining two points: The gradient of the line through (x 1, y 1 ) and (x 2, y 2 ) is: y 2 – y 1 x 2 – x 1 Perpendicular lines: If a line has gradient m, then a line perpendicular to this line has gradient 1m1m –