Normal Form or Perpendicular Form

1. Concept Overview

The normal form of a straight line expresses the equation of a line in terms of:

• The perpendicular distance of the line from the origin, and
• The angle that this perpendicular (normal) makes with the positive x-axis.

Instead of focusing on slope or intercepts, this form focuses on distance and direction.

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2. Statement of Normal Form

The equation of a straight line in normal form is:

[ x \cos α + y \sin α = p ]

where:

• [p] = perpendicular distance of the line from the origin
• [α] = angle between the perpendicular (normal) and the positive x-axis

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3. Geometrical Interpretation (Very Important)

• Draw a perpendicular from the origin to the line.
• Let its length be [p].
• Let the angle made by this perpendicular with the positive x-axis be [α].

Then every point [(x, y)] on the line satisfies:

[ x \cos α + y \sin α = p ]

Key Idea:
Normal form describes a line using its closest approach to the origin.

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4. Convert from General form to Normal Form

Given:
[ Ax + By + C = 0 ]

Divide by [\sqrt{A^2 + B^2}]:

[ \dfrac{A}{\sqrt{A^2 + B^2}} x + \dfrac{B}{\sqrt{A^2 + B^2}} y ][= −\dfrac{C}{\sqrt{A^2 + B^2}} ]

Comparing with:
[ x cos α + y sin α = p ]

We get:

• [cos α = \dfrac{A}{\sqrt{A^2 + B^2}}]
• [sin α = \dfrac{B}{\sqrt{A^2 + B^2}}]
• [p = \dfrac{|C|}{\sqrt{A^2 + B^2}}]

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5. Solved Examples

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Example 1. Convert the line [3x + 4y − 10 = 0] into normal form.

Step-by-Step Solution:

1. Compare with general form [Ax + By + C = 0]
   [A = 3], [B = 4], [C = −10]
2. Find [\sqrt{A^2 + B^2}]:
   [\sqrt{3^2 + 4^2} = \sqrt{25} = 5]
3. Divide entire equation by 5:
   [\dfrac{3}{5}x + \dfrac{4}{5}y = 2]
4. Compare with normal form [x cos α + y sin α = p]

Conclusion:
Normal form is:
[ x \dfrac{3}{5} + y \dfrac{4}{5} = 2 ]
where [p = 2], [cos α = \dfrac{3}{5}], [sin α = \dfrac{4}{5}].

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Example 2. Find the normal form of the line [x − y + 4 = 0].

Step-by-Step Solution:

1. Identify coefficients:
   [A = 1], [B = −1], [C = 4]
2. Compute:
   [\sqrt{A^2 + B^2} = \sqrt{1 + 1} = \sqrt{2}]
3. Divide the equation by [\sqrt{2}]:
   [\dfrac{x}{\sqrt{2}} − \dfrac{y}{\sqrt{2}} = −\dfrac{4}{\sqrt{2}}]
4. Make RHS positive (shift α by π):
   [ −\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = \dfrac{4}{\sqrt{2}} ]

Conclusion:
Normal form is:
[ −\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = 2\sqrt{2} ]

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Example 3. Write the normal form of the line [y = 6].

Step-by-Step Solution:

1. Convert to general form:
   [y − 6 = 0]
2. Identify coefficients:
   [A = 0], [B = 1], [C = −6]
3. [\sqrt{A^2 + B^2} = 1]
4. Divide entire equation by 1:
   [y = 6]
5. Compare with normal form.

Conclusion:
Normal form is:
[ x·0 + y·1 = 6 ]
with [p = 6], [α = 90°].

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Example 4. Find the perpendicular distance of the line [5x + 12y − 13 = 0] from the origin.

Step-by-Step Solution:

1. Use distance formula from origin:
   [p = \dfrac{|C|}{\sqrt{A^2 + B^2}}]
2. Substitute values:
   [p = \dfrac{13}{\sqrt{25 + 144}}]
3. Simplify:
   [p = \dfrac{13}{13} = 1]

Conclusion:
Perpendicular distance from origin is 1 unit.

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Example 5. Write the normal form of the line [x + y = 0].

Step-by-Step Solution:

1. Rewrite as:
   [x + y + 0 = 0]
2. [A = 1], [B = 1], [C = 0]
3. [\sqrt{A^2 + B^2} = \sqrt{2}]
4. Divide equation:
   [\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = 0]

Conclusion:
Normal form is:
[ x \dfrac{1}{\sqrt{2}} + y \dfrac{1}{\sqrt{2}} = 0 ]

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6. Important Restrictions (Exam Gold)

1. [p ≥ 0] (distance cannot be negative)
2. [0 ≤ α < 2π]
3. The sign of [p] is adjusted by choosing appropriate [α]

Hidden Concept:
If distance comes out negative, angle is shifted by π to keep [p] positive.

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7. Key Formulas to Remember

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8. Important Special Cases

(A) Line parallel to x-axis

Equation: [y = k]

Normal form:
[x·0 + y·1 = |k|]

So:

• [α = 90°]
• [p = |k|]

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(B) Line parallel to y-axis

Equation: [x = k]

Normal form:
[x·1 + y·0 = |k|]

So:

• [α = 0°]
• [p = |k|]

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9. Why Normal Form is Extremely Important in Exams

Normal form is mainly used for:

• Finding distance of a line from origin
• Finding distance between two parallel lines
• Understanding angle-based geometry
• Competitive exams (JEE, NDA, etc.)

CBSE Insight:
Distance-based questions almost always trace back to normal form.

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10. Common Conceptual Traps

• Taking [p] as signed (it is always non-negative)
• Confusing slope angle with normal angle
• Forgetting to divide by [\sqrt{A^2 + B^2}]
• Assuming α is the angle of the line (it is NOT)

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11. Comparison with Other Forms

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Practice Questions with Step-by-Step Solution

Question 1. Convert the line [2x + 2y − 4 = 0] into normal form.

Step-by-Step Solution:

Compare with general form [Ax + By + C = 0]
[A = 2], [B = 2], [C = −4]

Find [\sqrt{A^2 + B^2}]:
[\sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}]

Divide the entire equation by [2\sqrt{2}]:
[\dfrac{2x}{2\sqrt{2}} + \dfrac{2y}{2\sqrt{2}} = \dfrac{4}{2\sqrt{2}}]

Simplify each term:
[\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = \sqrt{2}]

Conclusion:

Normal form is
[ x \dfrac{1}{\sqrt{2}} + y \dfrac{1}{\sqrt{2}} = \sqrt{2} ]

Question 2. Find the perpendicular distance of the line [x − 3y + 6 = 0] from the origin.

Step-by-Step Solution:

Identify coefficients:
[A = 1], [B = −3], [C = 6]

Formula for perpendicular distance from origin:
[p = \dfrac{|C|}{\sqrt{A^2 + B^2}}]

Substitute values:
[p = \dfrac{6}{\sqrt{1^2 + (−3)^2}}]

Simplify denominator:
[p = \dfrac{6}{\sqrt{1 + 9}} = \dfrac{6}{\sqrt{10}}]

Conclusion:

Perpendicular distance =
[ \dfrac{6}{\sqrt{10}} ]

Question 3. Write the normal form of the line [3y − 4x = 0].

Step-by-Step Solution:

Rewrite in standard form:
[−4x + 3y + 0 = 0]

Identify coefficients:
[A = −4], [B = 3], [C = 0]

Find [\sqrt{A^2 + B^2}]:
[\sqrt{16 + 9} = \sqrt{25} = 5]

Divide entire equation by 5:
[−\dfrac{4}{5}x + \dfrac{3}{5}y = 0]

Conclusion:

Normal form is
[ −\dfrac{4}{5}x + \dfrac{3}{5}y = 0 ]

Question 4. Convert the line [y = 2x + 1] into normal form.

Step-by-Step Solution:

Convert into general form:
[2x − y + 1 = 0]

Identify coefficients:
[A = 2], [B = −1], [C = 1]

Compute [\sqrt{A^2 + B^2}]:
[\sqrt{4 + 1} = \sqrt{5}]

Divide entire equation by [\sqrt{5}]:
[\dfrac{2}{\sqrt{5}}x − \dfrac{1}{\sqrt{5}}y = −\dfrac{1}{\sqrt{5}}]

Conclusion:

Normal form is
[ x \dfrac{2}{\sqrt{5}} − y \dfrac{1}{\sqrt{5}} = −\dfrac{1}{\sqrt{5}} ]

Question 5. Find the value of [p] for the line [6x + 8y + 10 = 0].

Step-by-Step Solution:

Identify coefficients:
[A = 6], [B = 8], [C = 10]

Use distance formula:
[p = \dfrac{|C|}{\sqrt{A^2 + B^2}}]

Substitute values:
[p = \dfrac{10}{\sqrt{36 + 64}}]

Simplify:
[p = \dfrac{10}{\sqrt{100}} = \dfrac{10}{10}]

Conclusion:

[p = 1]

Question 6. Find the normal form of the line passing through the origin and making an angle 30° with the x-axis.

Step-by-Step Solution:

Slope of the line:
[m = \tan 30° = \dfrac{1}{\sqrt{3}}]

Equation through origin:
[y = \dfrac{1}{\sqrt{3}}x]

Convert to general form:
[\dfrac{1}{\sqrt{3}}x − y = 0]

Identify coefficients:
[A = \dfrac{1}{\sqrt{3}}], [B = −1]

Compute magnitude:
[\sqrt{\dfrac{1}{3} + 1} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}]

Divide entire equation by [\dfrac{2}{\sqrt{3}}]:
[\dfrac{1}{2}x − \dfrac{\sqrt{3}}{2}y = 0]

Conclusion:

Normal form is
[ x \dfrac{1}{2} − y \dfrac{\sqrt{3}}{2} = 0 ]

Question 7. Find the equation of the line whose normal makes an angle 45° with x-axis and distance from origin is 5.

Step-by-Step Solution:

Normal form equation:
[x cos α + y sin α = p]

Given:
[α = 45°], [p = 5]

Substitute trigonometric values:
[\cos 45° = \sin 45° = \dfrac{1}{\sqrt{2}}]

Final equation:
[\dfrac{x}{\sqrt{2}} + \dfrac{y}{\sqrt{2}} = 5]

Conclusion:

Required equation is
[ x \dfrac{1}{\sqrt{2}} + y \dfrac{1}{\sqrt{2}} = 5 ]

Question 8. Compare distances of the lines [3x − 4y + 12 = 0] and [6x − 8y + 12 = 0] from the origin.

Step-by-Step Solution:

Line 1:

[p_1 = \dfrac{12}{\sqrt{9 + 16}} = \dfrac{12}{5}]

Line 2:

[p_2 = \dfrac{12}{\sqrt{36 + 64}} = \dfrac{12}{10}]

Compare:
[\dfrac{12}{5} > \dfrac{12}{10}]

Conclusion:

First line is farther from the origin.

Question 9. If the normal form of a line is [x \cos α + y \sin α = 4], find its distance from origin.

Step-by-Step Solution:

Compare with standard normal form.

Distance from origin is [p].

Here, [p = 4].

Conclusion:

Distance from origin = 4 units.

Question 10. Can the normal form represent the line [x = 0]? Justify.

Step-by-Step Solution:

Equation [x = 0] implies the line passes through origin.

Distance from origin [p = 0].

Normal form requires [p ≥ 0].

Conclusion:

Yes, it can be represented as
[ x·1 + y·0 = 0 ]
which is a valid normal form.