# Core 1 - Coordinate Geometry Revision

Formulas to be memorised

- Created by: Mahek Jazmin
- Created on: 13-08-11 18:05

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

gradients is 1.

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.

## Comments