Coordinate Geometry

The equation of a straight line can be written as y = mx + c, where m is the gradient and c is the intercept with the vertical axis. Lines are parallel if they have the same gradient. Two lines are perpendicular if the product of their gradients is -1.

The gradient of a line passing through the points ( x1 , y1 ) and ( x2 , y2 )

y − y1 is 2 . x2 − x1

Coordinate Geometry

The distance between the points with coordinates

The midpoint of the line joining the points

( x1 , y1 ) and ( x2 , y2 )

( x1 , y1 ) and ( x2 , y2 )

is

( x2 − x1 )

2

+ ( y 2 − y1 ) . 2

Example: Find the equation of the perpendicular bisector of the line joining the points (3, 2) and (5, -6). Solution: The midpoint of the line joining (3, 2) and  3 + 5 2 + (−6) 

, , i.e. ( 4, −2 ) . (5, -6) is  2   2 The gradient of the line joining these two points

is:

−6 − 2 −8 = = −4 . 5−3 2

The gradient of the perpendicular bisector must therefore be −1 −4 = 1 4 . We need the equation of the line through (4, -2) with gradient ¼ . This is y − ( −2) = 14 ( x − 4) y + 2 = 14 x − 1 y = 14 x − 3

 x + x y + y2  is  1 2 , 1 . 2   2

Example: Find the point of intersection of the lines: 2x + y = 3  and y = 3x – 1.  Solution: To find the point of intersection we need to solve the equations  and  simultaneously. We can substitute  into equation : 2x + (3x – 1) = 3 i.e. 5x – 1 = 3 i.e. x = 4/5 Substituting this into equation : y = 3(4/5) – 1 = 7/5. Therefore the lines intersect at the point (4/5, 7/5).

The equation of the straight line with gradient m that passes through the point ( x1 , y1 ) is y − y1 = m( x − x1 ) . If the gradient of a line is m, then the gradient of a perpendicular line is −

1 . m

The equation of a circle centre (a, b) with radius r is ( x − a )2 + ( y − b) 2 = r 2 . Example: Find the centre and the radius of the circle with equation x 2 + 2 x + y 2 − 6 x + 6 = 0 . Solution: We begin by writing x 2 + 2 x in completed square form: 2 2 2 x 2 + 2 x = ( x + 1) − 1 = ( x + 1) − 1 . We then write y 2 − 6 x in completed square form: y 2 − 6 x = ( y − 3) 2 − 32 = ( y − 3) 2 − 9 . So we can rewrite the equation of the circle as ( x + 1) 2 − 1 + ( y − 3) 2 − 9 + 6 = 0 ( x + 1) 2 + ( y − 3) 2 = 4 . i.e. This is a circle centre (-1, 3), radius 2.