1. Coordinate Geometry II2 4 6 -2
X-axis
Y-axis
y = mx + b
ax + by + c = 0
Line Equations I
By Mr Porter

2. Assumed Knowledge.
It is assumed that you have completed the basic of the power-point
Coordinate Geometry I.
You are familiar with calculating the gradient of the line joining two points.
Test Yourself:
Find the gradient of the line joining the points P(-2,4) and Q(3,-2).

3. Equation of a Line through two Points: A(x1,y1) and B(x2, y2).
Select a general point on the line joining AB with coordinates P(x,y).
Now, the three points are collinear (lie in the same straight line).
Hence, the gradient of AB = gradient of AP = gradient of BP.
Two point formula of a line.
Example 1: Find the equation of the line joining A(1,2) and (4,5).
State the formula clearly:
, rearrange to make y the subject.
Substitute the values:
The equation of the required line is y = x + 1.
(Standard form)
simplify the R.H.S.
The equation can be written in standard form or general form: ax + by + c = 0

4. Example 2: Find the equation of a
line through P(-5,-2) and Q(6,4).
Example 3: Find the equation of a
line through H(7,-2) and K(-2,5).
State the formula clearly:
State the formula clearly:
Substitute the values :
Substitute the values :
simplify the R.H.S.
simplify the R.H.S.
, rearrange to
general form.
, rearrange to
general form.
11(y – 4) = 6(x – 6)
-9(y + 2) = 7(x – 7)
11y – 44 = 6x – 36
-9y – 18 = 7x – 49
0 = 6x – 11y +8
0 = 7x + 9y – 31
The required line is: 6x – 11y + 8 = 0 .
The required line is: 7x + 9y – 31 = 0 .

5. Exercise 1:
Find the equation of the line joining the following two points:
a) A(1,3) and B(5,11)
b) C(-1,-3) and D(3,5)
Ans: y = 2x + 1
Ans: y = 2x – 1
c) E(0,6) and F(3,-2)
d) G(-5,6) and B(4,-1)
Ans: 8x + 3y - 18 = 0
Ans: 7x + 9y - 19 = 0
Remember the formula:
2 Point Equation of a line.

6. Point - Gradient formula.
- This is a preferred method, used with DIFFERENTIATION.
Definition: The equation of a line with gradient , m, passing through a point P(x1, y1) is given by:
y – y1 = m (x – x1)
Example 1: Find the equation of a line
passing through A(1,3) and B(5,-3).
Example 2: Find the equation of a line
passing through P(-5,-3) and Q(3,4).
Gradient:
Gradient:
Equation: y – y1 = m (x – x1)
Equation: y – y1 = m (x – x1)
The required line is: 3x + 2y – 9 = 0 .
The required line is: 7x – 8y – 4 = 0 .

7. Exercise 2:
Find the equation of the line joining the following two points:
a) A(5,3) and B(-4,11)
b) C(-4,-2) and D(8,5)
Ans: 14x – 9y – 97 = 0
Ans: 7x – 12y +4 = 0
c) E(0,-2) and F(2,8)
d) G(-4,-6) and B(4,3)
Ans: y = 5x – 2
Ans: 9x – 8y - 12 = 0
Remember the formula:
Point-Gradient Equation of a line.

8. More Point - Gradient Lines.
Example 1: Find the equation of a line
with gradient m = 3 passing through
the point (1,7).
Example 2: Find the equation of a line
with gradient m = passing through
the point (-6,4).
Do you have a point? Yes, (1,6)
Do you have a point? Yes, (-6,4)
Do you have a gradient m? Yes , m = 3
Do you have a gradient m? Yes , m =
Equation: y – y1 = m (x – x1)
Equation: y – y1 = m (x – x1)
Substitute: y – 7 = 3 (x – 1)
Substitute: y – 4 = (x – -6)
Expand: y – 7 = 3x – 3
Expand: 2y – 8 = -3x – 18
Rearrange: y = 3x + 4
Rearrange: 3x + 2y + 10 = 0
The required line is: y = 3x + 4 .
The required line is: 3x + 2y + 10 = 0 .

9. Exercise 3:
Find the equation of the lines:
a) A(5,3) with gradient m = -3
b) B(-4,-2) with gradient m = 4
Ans: y = -3x + 18
Ans: y = 4x + 14
c) C(0,-2) with gradient m =
d) D(-4,-6) with gradient m =
Ans: 3x – 4y – 8 = 0
Ans: 5x + 3y + 38 = 0