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**

**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.