1. Co-ordinate Geometry III2 4 6 -2
X-axis
Y-axis
Parallel and Perpendicular Lines.
m1x m2 = -1
m1 = m2
Equations of Lines II
By Mr Porter

2. Assumed Knowledge: Gradient, m, using 2 points.
Standard form of a line: y = mx + b
General form of a line: Ax + By + C = 0
Ability to rearrange algebraic expression / equation: Change of Subject
Examples
Rearrange the general line Ax + By + C = 0, to standard form, y = mx + b.
Where m is the gradient and b is the y-intercept, for each of the following.
a) 3x + 2y – 6 = 0
b) 4x – 2y + 7 = 0
2y =-3x + 6
4x + 7 = 2y
y = mx + b
y = mx + b

3. Parallel Lines
Definition: Two lines y = m1x + b1 and y = m2 x + b2 are parallel, if their gradients are equal: m1 = m2.
Example 2
Find the equation of the line parallel to
, passing through the point (-1,3).
Example 1
Find the equation of the line parallel to y = 3x – 2, passing through the point (2,-5).
From, , in standard form,
y = mx + b.
To find the equation of a line, you need to have a gradient, m, and a point on the line. In this case, we have the point, need to find the gradient.
From, y =3x – 2, in standard form,
y = mx + b.
The gradient of the given line is
The gradient of the given line is m = 3.
Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2
Gradient, mi, of ALL lines parallel to the given line are equal. m1 = m2
Therefore, m = 3.
Therefore,
The point-gradient form of a lines is:
y - y1 = m(x - x1)
The point-gradient form of a lines is:
y - y1 = m(x - x1)
y – 3 = (x – -1)
y – -5 = 3(x – 2)
3y – 9 = -2x – 2
y + 5 = 3x – 6
y = 3x – 1, is the required line.
2x +3y – 7 = 0, is the required line.

4. Parallel lines to a given general line: Ax + By + C = 0.
If the given line is in general form, it has to be rearrange to standard form to obtain the gradient, m.
Example 1
Find the equation of the line parallel to 3x + 2y – 6 = 0, through the point (3,-5).
Example 2
Find the equation of the line parallel to 5x – 2y + 4 = 0, through the point (-2,8).
Step 1: Rearrange the given equation to standard form: y = mx + b.
Step 1: Rearrange the given equation to standard form: y = mx + b.
5x – 2y + 4 = 0
3x + 2y – 6 = 0
5x + 4 = 2y
2y = -3x + 6
i.e y = mx + b
i.e y = mx + b
The point-gradient form of a lines is:
y - y1 = m(x - x1)
The point-gradient form of a lines is:
y - y1 = m(x - x1)
3x + 2y + 1 = 0, is the required line.
5x – 2y + 26 = 0, is the required line.

5. Exercise 1: Find the equation of the line parallel to the given line passing through the given point.

6. Perpendicular Lines.
Definition: Two lines y = m1x + b1 and y = m2 x + b2 are perpendicular, if the  product of the gradients is equal to -1: m1 x m2 = -1.
Alternative: The gradients are negative reciprocals.
Example 1
Find the equation of the line perpendicular to y = 3x – 2, passing through the point (2,-5).
Example 2
Find the equation of the line perpendicular to , passing through the point (-1,3).
From, y =3x – 2, in standard form,
y = mx + b.
From, , in standard form,
y = mx + b.
The gradient of the given line is m1 = 3.
The gradient of the given line is
Perpendicular gradient, m2, are such that 3 x m2 = -1.
Therefore,
Perpendicular gradient, m2, are such that x m2 = -1.
Therefore,
Negative reciprocal!
Negative reciprocal!
The point-gradient form of a lines is:
y - y1 = m2(x - x1)
The point-gradient form of a lines is:
y - y1 = m2(x - x1)
x + 3y + 13 = 0, is the required line.
3x – 2y + 9 = 0, is the required line.

7. Perpendicular gradient, m2, are such that x m2 = -1. Therefore,
Example 3
Find the equation of the line perpendicular to 3x + 2y – 6 = 0, through the point (3,-5).
Example 4
Find the equation of the line perpendicular to 5x – 2y + 4 = 0, through the point (-2,8).
Step 1: Rearrange the given equation to standard form: y = mx + b.
Step 1: Rearrange the given equation to standard form: y = mx + b.
3x + 2y – 6 = 0
5x – 2y + 4 = 0
2y = -3x + 6
5x + 4 = 2y
i.e y = mx + b
i.e y = mx + b
Perpendicular gradient, m2, are such that x m2 = -1.
Therefore,
Perpendicular gradient, m2, are such that x m2 = -1.
Therefore,
The point-gradient form of a lines is:
y - y1 = m(x - x1)
The point-gradient form of a lines is:
y - y1 = m(x - x1)
2x – 3y – 21 = 0, is the required line.
2x + 5y – 36 = 0, is the required line.

8. Exercise 2: Find the equation of the line perpendicular to the given line passing through the given point.