Normal Form of Equation of Line Formula | Solved Examples | Practice Problems, FAQs, Important Questions

Introduction

The normal form of the equation of a line is a special way to represent a straight line using its perpendicular (normal) distance from the origin. Instead of using slope or intercepts, this form directly expresses the equation in terms of the shortest distance from the origin to the line and the angle made by the normal with the positive x-axis.

This method is particularly useful in coordinate geometry as it simplifies calculations involving distance, angles, and transformations of lines.

Equation of Line in Normal Form

Consider a perpendicular (ON) from the origin having length p to required line AB and it makes an angle α with the positive X-axis.

ON = p

∠AON = α

Let ON be perpendicular from the origin to line AB.

Therefore, Coordinates of the point N are :

N(p cosα , p sinα)

The slope of the line ON is tanα.

Therefore slope of line AB, which is perpendicular to line ON is (-1/tanα) because product of slopes of perpendicular lines is -1.

Hence slope of required line AB is (-1/tanα) = -cos α / sin α.

Hence, equation of the line AB having slope -cos α/sin α and passing through the point N(p cosα , p sinα) is,

y – p sinα = (-cos α/sin α)(x – p cosα)

After solving the above equation, we get equation of required line AB is

x cos α + y sin α = p

Equation of the line whose length of the perpendicular from the origin is p and the angle made by the perpendicular with the positive x-axis is given by α is given by:

x cos α + y sin α = p

This is known as the normal form of the line.

Note:    

(i) The equation of a, straight line in the form of x cos α + y sin α = p is called its normal form.

(ii) In equation x cos α + y sin α = p, the value of p is always positive and 0 ≤ α≤ 360°.

Transformation of general form of line into normal form

We will learn the transformation of general form into normal form.

To reduce the general equation Ax + By + C = 0 into normal form (x cos α + y sin α = p):

We have the general equation Ax + By + C = 0.

Let the normal form of the given equation ax + by + c = 0……………. (i) be  

x cos α + y sin α – p = 0, where p > 0. ……………. (ii)

Then, the equations (i) and (ii) are the same straight line i.e., identical.

⇒ \(\frac{A}{cos α}\) = \(\frac{B}{sin α}\) = \(\frac{C}{-p}\)

⇒ \(\frac{C}{P}\) = \(\frac{-A}{cos α}\) = \(\frac{-B}{sin α}\) = \(\frac{+\sqrt{a^{2} + b^{2}}}{\sqrt{cos^{2} α + sin^{2} α}}\) = +  \(\sqrt{A^{2} + B^{2}}\)

Therefore, p = \(\frac{C}{\sqrt{A^{2} + B^{2}}}\), cos α = – \(\frac{A}{\sqrt{A^{2} + B^{2}}}\) and sin α = – \(\frac{B}{\sqrt{A^{2} + B^{2}}}\)

So, putting
the values of cos α, sin α and p in the equation (ii) we get the form,

⇒ –
\(\frac{A}{\sqrt{A^{2} + B^{2}}}\) x – \(\frac{B}{\sqrt{A^{2} + B^{2}}}\) y – \(\frac{C}{\sqrt{A^{2}
+ B^{2}}}\) =  0, when c > 0

⇒ \(\frac{A}{\sqrt{A^{2}
+ B^{2}}}\) x +  \(\frac{B}{\sqrt{A^{2} +
B^{2}}}\) y = – \(\frac{C}{\sqrt{A^{2} + B^{2}}}\), when c < 0

Which is the required normal form of the general form of equation Ax + By + C = 0.

Solved Examples

Example.1 : Find the equation of the straight line which is at a of distance 7 units from the origin and the perpendicular from the origin to the line makes an angle 45° with the positive direction of x-axis.

Solution:

We know that the equation of the straight line upon which the length of the perpendicular from the origin is p and this perpendicular makes an angle α with x-axis is x cos α + y sin α = p.

Here p = 7 and α = 45°

Therefore, the equation of the straight line in normal form is

x cos 45° + y sin 45° = 7

⇒ x ∙ 1√2 + y ∙ 1√2 = 7

⇒ x + y = 7√2, which is the required equation.

Examples 2: Convert the given equation in the normal form 2x – 2y – 6 = 0.

Solution:

Given: 2x – 2y – 6 = 0

Divide the given equation

√(2)² + (-2)² = √4 + 4 = √8 = 2√2

So, 2x/2√2 – 2y/2√2 = 6/2√2

x/√2 – y/√2 = 3/√2…(1)

As we know that the Intercept Form is 

xcosα + ysinα = p…(2)

On comparing eq(1) and (2) we get

cosα = 1/√2

sinα = -1/√2

So, xcos315° + ysin315° = 3/√2

Example 3: Find the value of p and α, equation is x + y + 3 = 0.

Solution:

Given: x + y + 3 = 0

Divide the given equation

√(1)² + (1)² = √2

So, x/√2 + y/√2 = -3/√2…(1)

As we know that the Intercept Form is 

xcosα + ysinα = p…(2)

On comparing eq(1) and (2) we get

cosα = 1/√2

sinα = 1/√2

xcos45° + ysin45° = -3/√2

Hence, p = -3/√2 and α = 45°

Example 4: Reduce the line 4x + 3y – 19 = 0 to the normal form.

Solution:

The given equation is 4x + 3y – 19 = 0

First shift the constant term (-19) on the RHS and make it positive.

4x + 3y = 19 ………….. (i)

Now determine \(\sqrt{(\textrm{Coefficient of x})^{2} +(\textrm{Coefficient of y})^{2}}\)

= \(\sqrt{(4)^{2} + (3)^{2}}\)

= \(\sqrt{16 + 9}\) = √25 = 5

Now dividing both sides of the equation (i) by 5, we get

\(\frac{4}{5}\)x + \(\frac{3}{5}\)y = \(\frac{19}{5}\)

Which is the normal form of the given equation 4x + 3y – 19 = 0.

Example 5: Transform the equation 3x + 4y = 5√2 to normal form and find the perpendicular distance from the origin of the straight line; also find the angle that the perpendicular makes with the positive direction of the x-axis.

Solution:    

The given equation is 3x + 4y = 5√2 ……..….. (i)

Dividing both sides of equation (1) by + \(\sqrt{(3)^{2} + (4)^{2}}\) = +
5 we get,

⇒ \(\frac{3}{5}\)x + \(\frac{4}{5}\)y = \(\frac{5√2}{5}\)

⇒ \(\frac{3}{5}\)x + \(\frac{4}{5}\)y = √2

Which is the normal form of the given equation 3x + 4y = 5√2.

Therefore, the required, perpendicular distance from the origin
of the straight line (i) is √2 units.

If the perpendicular makes an angle α with the positive direction of the x-axis then,

cos α = \(\frac{3}{4}\) and sin α = \(\frac{4}{5}\)

Therefore,
tan α = \(\frac{sin α}{cos α }\) = \(\frac{\frac{4}{5}}{\frac{3}{5}}\) = \(\frac{4}{3}\)

⇒ α = tan\(^{-1}\)\(\frac{4}{3}\).

Key Features of the Normal Form

• The equation represents a line in terms of the shortest distance from the origin.
• It is independent of the conventional slope-based representation.
• The angle α\alphaα helps in determining the orientation of the line in the coordinate plane.
• It is useful in problems involving perpendicular distances and angles between lines.

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Applications of the Normal Form

1. Finding the Distance of a Line from the Origin
   • The normal form provides a direct way to calculate the shortest distance of a given line from the origin without additional transformations.
2. Determining Parallel and Perpendicular Lines
   • Lines with the same normal distance ppp but different angles α\alphaα can be easily compared to identify whether they are parallel or perpendicular.
3. Geometric Transformations
   • When rotating a line around the origin, the normal form makes it easier to compute new equations without recalculating slopes.
4. Problem Solving in JEE and CBSE Board Exams
   • Many competitive exams include problems based on the normal form, especially in questions involving distances, angles, and geometric interpretations of lines.

Summary

In coordinate geometry, a straight line can be represented in different forms such as the slope-intercept form, two-point form, and general form. However, the normal form provides an alternative way to express the equation by using the perpendicular distance from the origin, making it an efficient tool in many geometric problems.

To define a line using the normal form, two key parameters are required:

1. The perpendicular distance from the origin to the line, which is denoted as ppp.
2. The angle made by the normal with the positive x-axis, which is denoted as α\alphaα.

This approach avoids the direct use of slope and provides a geometric perspective to line equations.

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Conclusion

The normal form of the equation of a line is an essential representation in coordinate geometry. It provides a unique way of defining a line using its perpendicular distance from the origin and the angle of inclination of the normal. This form is particularly useful in distance-based calculations, transformations, and understanding geometric properties of lines.

For more detailed explanations and solved examples, students preparing for CBSE Board Exams and JEE Mains & Advanced can download the full PDF from ANAND CLASSES. ✅