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Equations of Perpendicular Lines

In the diagram below, two lines l1 and l2 are perpendicular to each other i.e. PA ortho PB.

Let the gradient of PA be m1 and the gradient of PB be m2equation_of_perpendicular_lines

LABP = 180° – θ2

tan θ1 = m1 = {PB}/{AP}

tan (180 – θ2) = m2 = {AP}/{PB}  =  1/{tan θ_1}

Now tan 180 – θ2  =  -tan θ2

Therefore tan 180 – θ2  =  -tan θ2  =  1/{tan θ_1}

But tan θ1 = m1 (gradient of l1), and

tan θ2 = m2 (gradient of l2)

Hence -tan θ2 =  m2  =  1/{tan θ_1}  = 1/{m_1}

-m2  =  1/{m_1}

-m1m2  =  1, or

m1m2  =  -1

If two lines are perpendicular and have gradients m1 and m2 respectively, then

m1m2  =  -1  or  m1  =  {-1}/{m_2}

 

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

y = {5/3}~-~{2/3}x

m1-{2/3}

For line 2

3x – 2y + 6 = 0

2y = 3x + 6

y = {3/2}~+~{6/2}

y = {3/2} + 3

m23/2

Now m1 x m2  =  – {2/3}~*~{3/2}

=  -1

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.