We learnt from the equations of perpendicular lines that given two lines with gradients m_{1} and m_{2} respectively, the two lines will be perpendicular when the product of their gradients is equal to -1. That is –

m_{1}m_{2 } = -1 or m_{1} =

We now look at a few examples of equations of perpendicular lines, and how to solve them.

**Example 1**: Find the equation of a line that is perpendicular to 2x – 5y + 7 = 0 and passing through the point (2, 3)

We can solve this by two methods.

**Method 1**: First find the gradient of the given line.

2x – 5y + 7 = 0

2x + 7 = 5y

y =

So m_{1} =

Next find the gradient of the line perpendicular to 2x – 5y + 7 = 0.

m_{2} =

=

=

Given a point (2, 3) and gradient m_{2} = , use the point gradient form to find the equation of the straight line.

Equation of the required line is –

y – y_{1} = m(x – x_{1})

y – 3 = (x – 2)

Multiplying by 2 on both sides –

2y – 6 = -5(x – 2)

2y – 6 = -5x + 10

5x + 2y – 6 – 10 = 0

5x + 2y – 16 = 0 is the equation of the line perpendicular to 2x – 5y + 7 = 0 and passing through the point (2, 3).

**Method 2**: Using perpendicular property of lines, the line perpendicular to 2x – 5y + 7 = 0 is 5x + 2y + k = 0 (interchange of the x and y coefficients and change only one of the coefficients to the opposite sign).

Substitute (2, 3) in this equation and solve for k. So

5x + 2y + k = 0

5 x 2 + 2 x 3 + k = 0

10 + 6 + k = 0

k = -16

So the required equation of the line perpendicular to 2x – 5y + 7 = 0 and passing through (2, 3) is 5x + 2y – 16 = 0

** Example 2**: Show that the line ax + by + c = 0 is

- parallel to ax + by + d = 0, and
- perpendicular to bx – ay + e = 0

Given Line 1 to be ax + by + c = 0.

Gradient of Line 1 is –

by = – c – ax

y =

m_{1} = **Equation 1**

Gradient of Line 2 is –

by = – d – ax

y =

m_{2} = **Equation 2**

Gradient of Line 3 is –

ay = bx + e

y =

m_{3} = **Equation 3
**

From **1** and **2**, m_{1} = m_{2} =

Hence lines 1 and 2 are parallel.

From **1** and **3**, m_{1} = =

=

m_{1} =

Or m_{1} x m_{3} = -1.

Hence lines 1 and 3 are perpendicular to each other.

**Example 3**: A line is parallel to 4x – 5y + 1 = 0 and passes through the points (1, 3) and (5, a). Find the value of ‘a’.

Gradient of the line 4x – 5y + 1 = 0 is

5y = 4x + 1

y =

m_{1} = .

The line parallel to 4x – 5y +1 = 0 also has a gradient of .

Since the line passes through (1, 3) and (5, a), the gradient of the line passing through these points is also . So

gradient = m =

=

=

5(a – 3) = 4 x 4

5a – 15 = 16

5a = 31

a =

So the value of a =