1. Coordinate geometry
© Christine Crisp

2. m is the gradient of the line
The equation of a straight line is
m is the gradient of the line
c is the point where the line meets the y-axis,
the y-intercept
e.g has gradient m =
and y-intercept, c =
gradient = 2 x
intercept on y-axis

3. e.g. Substituting x = 4 in gives
gradient = 2 x
intercept on y-axis
( 4, 7 ) x
The coordinates of any point lying on the line satisfy the equation of the line
e.g. Substituting x = 4 in gives
showing that the point ( 4,7 ) lies on the line.

4. Finding the equation of a straight line when we know
its gradient, m and
the coordinates of a point on the line.
Using , m is given, so we can find c by
substituting for y, m and x.
e.g. Find the equation of the line with gradient
passing through the point
Solution:
(-1, 3) x
Notice that to find c, the equation has been solved from right to left. This takes a bit of practice but reduces the chance of errors.
So,

5. To find the equation of a straight line given 2 points on the line.
e.g. Find the equation of the line through the points
Solution: First find the gradient:
Now
on the line:
Equation of line is

6. We sometimes rearrange the equation of a straight line so that zero is on the right-hand side ( r.h.s. )
e.g can be written as
We must take care with the equation in this form.
e.g. Find the gradient of the line with equation
Solution: Rearranging to the form :
so the gradient is

7. Parallel and Perpendicular Lines
If 2 lines have gradients and , then:
They are parallel if
They are perpendicular if

8. e.g. 1 Find the equation of the line parallel to
which passes through the point
Solution: The given line has gradient 2. Let
For parallel lines,
is the equation of any line parallel to
Using
on the line

9. e.g. Find the equation of the line perpendicular to
passing through the point
Solution: The given line has gradient 2. Let
Perpendicular lines:
Equation of a straight line:
on the line
We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:

10. SUMMARY
Method of finding the equation of a straight line:
If the gradient isn’t given, find the gradient using
either parallel lines:
or perpendicular lines:
or 2 points on the line:
Substitute for y, m and x in into
to find c.

11. A Second Formula for a Straight Line ( optional )
Let ( x, y ) be any point on the line x x
Let be a fixed point on the line

12. To use the formula we need to be given
either: one point on the line and the gradient
or: two points on the line
e.g. Find the equation of the line through the points
Solution: First find the gradient
Now use with
We could use the 2nd point,
(-1, 3) instead of (2, -3)

13. The mid-point is the average of the end points:
M X
The mid-point is the average of the end points:

14. A formula for the Mid-Point of AB
M X
The mid-point is the average of the end points: or

15. Exercise
Find the mid-point, M of the line joining to
Solution:

16. Using Pythagoras’ theorem:

17. A formula for the length of the line joining A to B
Using Pythagoras’ theorem:

19. The following slides contain repeats of information on earlier slides, shown without colour, so that they can be printed and photocopied.
For most purposes the slides can be printed as “Handouts” with up to 6 slides per sheet.

20. If 2 lines have gradients and , then:
They are parallel if
They are perpendicular if
If 2 lines have gradients and , then:
Equation of a straight line
Gradient of a straight line
where and are points on the line
where m is the gradient and c is the intercept on the y-axis
SUMMARY

21. Solution: First find the gradient:
e.g. Find the equation of the line through the points
Now
on the line:
Equation of line is

22. We don’t usually leave fractions ( or decimals ) in equations
We don’t usually leave fractions ( or decimals ) in equations. So, multiplying by 2:
e.g. Find the equation of the line perpendicular to
passing through the point
Solution: The given line has gradient 2. Let
Perpendicular lines:
Equation of a straight line:
on the line