In the diagram below, two lines l1 and l2 are perpendicular to each other i.e. PA PB.
LABP = 180° – θ2
tan θ1 = m1 =
tan (180 – θ2) = m2 = =
Now tan 180 – θ2 = -tan θ2
Therefore tan 180 – θ2 = -tan θ2 =
But tan θ1 = m1 (gradient of l1), and
tan θ2 = m2 (gradient of l2)
Hence -tan θ2 = m2 = =
-m1m2 = 1, or
m1m2 = -1
If two lines are perpendicular and have gradients m1 and m2 respectively, then
m1m2 = -1 or m1 =
Example 1: Show that the lines 2x + 3y – 5 = 0 and 3x – 2y + 6 = 0 are perpendicular to each other.
For line 1
2x + 3y – 5 = 0
For line 2
3x – 2y + 6 = 0
2y = 3x + 6
y = + 3
Now m1 x m2 = –
Hence the two lines are perpendicular to each other.
Note: In the above example, the two equations of perpendicular lines viz. 2x + 3y = 5 and 3x – 2y = -6 have the coefficients of x and y interchanged, and one of the coefficients is negative.
Hence we can state that given a line with equation ax + by + c1 = 0, the equation of the line perpendicular to it is given by bx – ay + c2 = 0
Equation of the line perpendicular to 3x – 5y + 6 = 0 will be given by 5x + 3y + k = 0 where k can be any real value satisfying a given condition.
Here are a few more examples of equations of perpendicular lines.