Straight Line in Normal Form

We will learn how to find the equation of a straight line in normal form.

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

If the line length of the perpendicular draw from the origin upon a line and the angle that the perpendicular makes with the positive direction of x-axis be given then to find the equation of the line.

Suppose the line AB intersects the x-axis at A and the y-axis at B. Now from the origin O draw OD perpendicular to AB.

Straight Line in Normal Form

The length of the perpendicular OD from the origin = p and ∠XOD = α, (0 ≤ α ≤ 2π).

Now we have to find the equation of the straight line AB.

Now, from the right-angled ∆ODA we get,

\(\frac{OD}{OA}\) = cos α        

⇒ \(\frac{p}{OA}\) = cos α          

⇒ OA = \(\frac{p}{cos α}\)

Again, from the right-angled ∆ODB we get,

∠OBD = \(\frac{π}{2}\) - ∠BOD = ∠DOX = α    

Therefore, \(\frac{OD}{OB}\) = sin α

or, \(\frac{p}{OB}\) = sin α     

or, OB = \(\frac{p}{sin α}\)

Since the intercepts of the line AB on x-axis and y-axis are OA and OB respectively, hence the required

\(\frac{x}{OA}\) + \(\frac{y}{OB}\) = 1        

⇒ \(\frac{x}{\frac{p}{cos α}}\) + \(\frac{y}{\frac{p}{sin α}}\) = 1

⇒ \(\frac{x cos α}{p}\) + \(\frac{y sin α}{p}\) = 1           

⇒ x cos α + y sin α = p, which is the required form.

Solved examples to find the equation of a straight line in normal form:

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 ∙ \(\frac{1}{√2}\) + y ∙ \(\frac{1}{√2}\) = 7

⇒ \(\frac{x}{√2}\) + \(\frac{y}{√2}\) = 7

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

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°.

● 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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