COORDINATE GEOMETRY
Week commencing Monday 2nd November 2009
Learning Intention:
Given a point and the gradient to be able to find the
equation of a line.
Contents:
1.
2.
3.
Equation of a Straight Line from gradient and a point
Examples
Assignment 2
COORDINATE GEOMETRY
EQUATION OF A LINE FROM POINT & GRADIENT
If given the gradient of a line and a point, (x1, y1), on
the line we can find the equation of line using the
formula:
y – y1 = m(x – x1)
COORDINATE GEOMETRY
EQUATION OF A LINE FROM POINT & GRADIENT
Example:
Find the equation of the line with gradient 4 that
passes through the point (1, 3).
Solution:
m = 4
(x1, y1) = (1, 3)
Substituting into formula y  y1  m(x  x1 ) we get:
y – 3 = 4(x – 1)
y – 3 = 4x – 4
expanding brackets
y = 4x – 4 + 3
simplifying
y = 4x – 1
in form y = mx + c
COORDINATE GEOMETRY
EQUATION OF A LINE FROM POINT & GRADIENT
Example:
Find the equation of the line with gradient -½ that
passes through the point (5, 3).
Solution:
m = -½
(x1, y1) = (5, 3)
Substituting into the formula we get:
1
y  3   (x  5)
2
1
5
y3   x
2
2
1
5
y   x 3
2
2
1
11
y x
2
2
COORDINATE GEOMETRY
EQUATION OF A LINE FROM POINT & GRADIENT
Example:
The line y = 4x – 8 meets the x-axis at the point A. Find
the equation of the line with gradient 3 that passes through
the point A.
Solution:
We first need a point on the required line. We can get this
by finding the coordinates of A.
Let y = 0 to find where y = 4x – 8 meets the x –axis.
4x – 8 = 0
4x = 8
x = 2
coordinates of A are (2, 0)
m = 3
(2, 0)
y – 0 = 3(x – 2)
y = 3x - 6
Substitute into formula
COORDINATE GEOMETRY
Assignment 2
Follow the link for Assignment 2 in the Moodle Course
Area underneath Coordinate Geometry. This is a
Yacapaca Activity.
Assignments to be completed by 5:00pm on Monday 9th
November 2009