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






11 and 12 Grade Math

From Equation of a Line Perpendicular to a Line to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Successor and Predecessor | Successor of a Whole Number | Predecessor

    May 24, 24 06:42 PM

    Successor and Predecessor of a Whole Number
    The number that comes just before a number is called the predecessor. So, the predecessor of a given number is 1 less than the given number. Successor of a given number is 1 more than the given number…

    Read More

  2. Counting Natural Numbers | Definition of Natural Numbers | Counting

    May 24, 24 06:23 PM

    Natural numbers are all the numbers from 1 onwards, i.e., 1, 2, 3, 4, 5, …... and are used for counting. We know since our childhood we are using numbers 1, 2, 3, 4, 5, 6, ………..

    Read More

  3. Whole Numbers | Definition of Whole Numbers | Smallest Whole Number

    May 24, 24 06:22 PM

    The whole numbers are the counting numbers including 0. We have seen that the numbers 1, 2, 3, 4, 5, 6……. etc. are natural numbers. These natural numbers along with the number zero

    Read More

  4. Math Questions Answers | Solved Math Questions and Answers | Free Math

    May 24, 24 05:37 PM

    Math Questions Answers
    In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students

    Read More

  5. Estimating Sum and Difference | Reasonable Estimate | Procedure | Math

    May 24, 24 05:09 PM

    Estimating Sum or Difference
    The procedure of estimating sum and difference are in the following examples. Example 1: Estimate the sum 5290 + 17986 by estimating the numbers to their nearest (i) hundreds (ii) thousands.

    Read More