Equation of a Line Perpendicular to a Line

We will learn how to find the equation of a line perpendicular to a line.

Prove that the equation of a line perpendicular to a given line ax + by + c = 0 is bx - ay + λ = 0, where λ is a constant.

Let m\(_{1}\) be the slope of the given line ax + by + c = 0 and m\(_{2}\) be the slope of a line perpendicular to the given line.

Then,

m\(_{1}\) = -\(\frac{a}{b}\) and m\(_{1}\)m\(_{2}\) = -1

⇒ m\(_{2}\) = -\(\frac{1}{m_{1}}\) = \(\frac{b}{a}\)

Let c\(_{2}\) be the y-intercept of the required line. Then its equation is

y = m\(_{2}\)x + c\(_{2}\)

⇒ y = \(\frac{b}{a}\) x + c\(_{2}\)

⇒ bx - ay + ac\(_{2}\) = 0

⇒ bx - ay + λ = 0, where  λ = ac\(_{2}\) = constant.

To get it more clear let us assume that ax + by + c = 0 (b ≠ 0) be the equation of the given straight line.

Now convert the ax + by + c = 0 in to slope-intercept form we get,

by = - ax - c     

⇒ y = - \(\frac{a}{b}\) x - \(\frac{c}{b}\)

Therefore, the slope of the straight line ax + by + c = 0 is (- \(\frac{a}{b}\)).

Let m be the slope of a line which is perpendicular to the line ax + by + c = 0. Then, we must have,

m × (- \(\frac{a}{b}\)) = - 1    

⇒ m = \(\frac{b}{a}\) 

Therefore, the equation of a line perpendicular to the line ax + by + c = 0 is

y = mx + c    

⇒ y = \(\frac{b}{a}\) x + c

⇒ ay =  bx +  ac

⇒ bx - ay+ k = 0, where k = ac, is an arbitrary constant.

 

Algorithm for directly writing the equation of a straight line perpendicular to a given straight line:

To write a straight line perpendicular to a given straight line we proceed as follows:

Step I: Interchange the coefficients of x and y in equation ax + by + c = 0.

Step II: Alter the sign between the terms in x and y of equation i.e., If the coefficient of x and y in the given equation are of the same signs make them of opposite signs and if the coefficient of x and y in the given equation are of the opposite signs make them of the same sign.

Step III: Replace the given constant of equation ax + by + c = 0 by an arbitrary constant.

For example, the equation of a line perpendicular to the line 7x + 2y + 5 = 0 is 2x - 7y + c = 0; again, the equation of a line, perpendicular to the line 9x - 3y = 1 is 3x + 9y + k = 0.

 

Note:

Assigning different values to k in bx - ay + k = 0 we shall get different straight lines each of which is perpendicular to the line ax + by + c = 0. Thus we can have a family of straight lines perpendicular to a given straight line.

Solved examples to find the equations of straight lines perpendicular to a given straight line

1. Find the equation of a straight line that passes through the point (-2, 3) and perpendicular to the straight line 2x + 4y + 7 = 0.

Solution:

The equation of a line perpendicular to 2x + 4y + 7 = 0 is

4x - 2y + k = 0 …………………… (i) Where k is an arbitrary constant.

According to the problem equation of the perpendicular line 4x - 2y + k = 0 passes through the point (-2, 3)

Then,

4 ∙ (-2) - 2 ∙ (3) + k = 0

⇒ -8 - 6 + k = 0

⇒ - 14 + k = 0

⇒ k = 14

Now putting the value of k = 14in (i) we get, 4x - 2y + 14 = 0

Therefore the required equation is 4x - 2y + 14 = 0.

2. Find the equation of the straight line which passes through the point of intersection of the straight lines x + y + 9 = 0 and 3x - 2y + 2 = 0 and is perpendicular to the line 4x + 5y + 1 = 0.

Solution:  

The given two equations are x + y + 9 = 0 …………………… (i) and 3x - 2y + 2 = 0 …………………… (ii)

Multiplying equation (i) by 2 and equation (ii) by 1 we get

                                                       2x + 2y + 18 = 0

                                                       3x  - 2y +   2 = 0

Adding the above two equations we get, 5x = - 20

⇒ x = - 4

Putting x = -4 in (i) we get, y = -5

Therefore, the co-ordinates of the point of intersection of the lines (i) and (ii) are (- 4, - 5).

Since the required straight line is perpendicular to the line 4x + 5y + 1 = 0, hence we assume the equation of the required line as

5x - 4y + λ = 0 …………………… (iii)

Where λ is an arbitrary constant.

By problem, the line (iii) passes through the point (- 4, - 5); hence we must have,

⇒ 5 ∙ (- 4) - 4 ∙ (- 5) + λ = 0  

⇒ -20 + 20 + λ = 0  

⇒ λ = 0.

Therefore, the equation of the required straight line is 5x - 4y = 0.

● The Straight Line

• Straight Line
• Slope of a Straight Line
• Slope of a Line through Two Given Points
• Collinearity of Three Points
• Equation of a Line Parallel to x-axis
• Equation of a Line Parallel to y-axis
• Slope-intercept Form
• Point-slope Form
• Straight line in Two-point Form
• Straight Line in Intercept Form
• Straight Line in Normal Form
• General Form into Slope-intercept Form
• General Form into Intercept Form
• General Form into Normal Form
  
• Point of Intersection of Two Lines
  
• Concurrency of Three Lines
• Angle between Two Straight Lines
• Condition of Parallelism of Lines
• Equation of a Line Parallel to a Line
• Condition of Perpendicularity of Two Lines
• Equation of a Line Perpendicular to a Line
• Identical Straight Lines
• Position of a Point Relative to a Line
• Distance of a Point from a Straight Line
• Equations of the Bisectors of the Angles between Two Straight Lines
• Bisector of the Angle which Contains the Origin
• Straight Line Formulae
• Problems on Straight Lines
• Word Problems on Straight Lines
• Problems on Slope and Intercept

11 and 12 Grade Math

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