Core 1 - Coordinate Geometry Revision

Formulas to be memorised

HideShow resource information

First 308 words of the document:

The equation of a straight line can be written as y =The gradient of a line passing through the points The equation of the straight line with gradient m that
mx + c, where m is the gradient and c is the
passes through the point is
intercept with the vertical axis.
.
.
Lines are parallel if they have the same gradient. If the gradient of a line is m, then the gradient of a
Two lines are perpendicular if the product of their
Coordinate Geometry
perpendicular line is .
The distance between the points with coordinates The midpoint of the line joining the points The equation of a circle centre (a, b) with radius r
. is .
.
Example: Example: Example:
Find the equation of the perpendicular bisector of Find the point of intersection of the lines: Find the centre and the radius of the circle with
the line joining the points (3, 2) and (5, 6). 2x + y = 3
equation .
and y = 3x ­ 1.
Solution:
Solution:
The midpoint of the line joining (3, 2) and Solution:
To find the point of intersection we need to solve We begin by writing in completed square
the equations and simultaneously. form:
(5, 6) is .
We can substitute into equation : = .
The gradient of the line joining these two points is: 2x + (3x ­ 1) = 3
i.e. 5x ­ 1 = 3 We then write in completed square form:
. i.e. x = 4/5 = .
The equation of the perpendicular bisector must Substituting this into equation :
So we can rewrite the equation of the circle as
y = 3(4/5) ­ 1 = 7/5.
therefore be . Therefore the lines intersect at the point
We need the equation of the line through (4, 2) (4/5, 7/5).
with gradient ¼ . This is i.e. .
This is a circle centre (1, 3), radius 2.