Condition of Perpendicularity

We will discuss here about the condition of perpendicularity of two straight lines.

Let the lines AB and CD be perpendicular to each other. If the inclination of AB with the positive direction of the x-axis is θ then the inclination of CD with the positive direction of the x-axis will be 90° + θ.

Therefore, the slope of AB = tan θ, and

the slope of CD = tan (90° + θ).

From trigonometry, we have, tan (90° + θ) = - cot θ

Therefore, if the slope of AB is m\(_{1}\) and

the slope CD = m\(_{2}\) then 

m\(_{1}\) = tan θ and m\(_{2}\) = - cot θ.

So, m\(_{1}\) ∙ m\(_{2}\) = tan θ ∙ (- cot θ) = -1

Two lines with slopes m\(_{1}\) and m\(_{2}\) are perpendicular to each other if and only if m\(_{1}\) ∙ m\(_{2}\) = -1

Note: (i) By the definition, the x-axis is perpendicular to the y-axis.

(ii) By definition, any line parallel to the x-axis is perpendicular to any line parallel to the y-axis.

(iii) If the slope of a line is m then any line perpendicular to it will have the slope \(\frac{-1}{m}\) (i.e., negative reciprocal of m).


Solved example on Condition of perpendicularity of two lines:

Find the equation of the line passing through the point (-2, 0) and perpendicular to the line 4x – 3y = 2.


First we need to express the given equation in the form y = mx + c.

Given equation is 4x – 3y = 2.

-3y = -4x + 2

y = \(\frac{4}{3}\)x - \(\frac{2}{3}\)

Therefore, the slope (m) of the given line = \(\frac{4}{3}\)

Let the slope of the required line be m\(_{1}\).

According to the problem the required line is perpendicular to the given line.

Therefore, from the condition of perpendicularity we get,

m\(_{1}\) ∙ \(\frac{4}{3}\) = -1

⟹ m\(_{1}\) = -\(\frac{3}{4}\)

Thus, the required line has the slope -\(\frac{3}{4}\) and it passes through the point (-2, 0).

Therefore, using the point-slope form we get

y - 0 = -\(\frac{3}{4}\){x - (-2)}

⟹ y = -\(\frac{3}{4}\)(x + 2)

⟹ 4y = -3(x + 2)

⟹ 4y = -3x + 6

⟹ 3x + 4y + 6 = 0, which is the required equation.

10th Grade Math

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