1. 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.
The gradient of a line passing through the
points ( x1 , y1 ) and ( x2 , y2 )
y − y1
is 2
.
x2 − x1
Lines are parallel if they have the same
gradient.
Two lines are perpendicular if the product of
their gradients is -1.
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
+ ( y2 − 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) 
,
(5, -6) is 
÷, i.e. ( 4, −2 ) .
2
 2

The gradient of the line joining these two points
is:
−6 − 2 −8
=
= −4 .
5−3
2
The equation 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) = 1 ( x − 4)
4
y + 2 = 1 x −1
4
y = 1 x−3
4
Coordinate Geometry
 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.