Math, asked by XxAashuxX, 3 months ago

Derive the equation of the line in normal form in terms of coordinate.

Chapter → Straight lines

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Answers

Answered by Anonymous
131

To prove :

  • The equation of the line in normal form

Proof :

Let " l " be the line, AB is perpendicular to x - axis and ∠AOB = ω

  • Perpendicular distance from origin be OA = p

As we know that

Slope of perpendicular of two lines

  • m₁m₂ = - 1

★ m₁ (l)

★ m₂ (OA)

→ Slope of line l and OA = - 1

→ m₁m₂ = - 1

→ m₁ = -1/m₂

→ m₁ = -1/tanω [ m = tan ∅ ]

→ m₁ = -1/sinω/cosω

→ m₁ = - cosω/sinω

  • Slope of line l = - cosω/sinω

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Now in triangle ∆OAB

  • sinω = AB/OA

sinω = AB/p

p sinω = AB

  • cosω = OB/OA

cosω = OB/p

p cosω = OB

  • Coordinate of A = (p cosω, p sinω)

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°•° We know that the equation of line in one point form is given by

  • y - y₀/x - x₀ = m

→ y - y₀ = m(x - x₀)

Substitute the values of coordinate A

→ y - p sinω = m(x - p cosω)

  • Put the value of slope of line l

→ y - p sinω = - cosω/sinω(x - p cosω)

→ sinω(y - p sinω) = - cosω(x - p cosω)

→ y sinω - p sin²ω = - x cosω + p cos²ω

→ y sinω + x cosω = p sin²ω + p cos²ω

→ y sinω + x cosω = p(sin²ω + cos²ω)

→ y sinω + x cosω = p [sin²∅ + cos²∅ = 1 ]

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Answered by Anonymous
3

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.

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,

ODOA = cos α

⇒ pOA = cos α

⇒ OA = pcosα

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

∠OBD = π2 - ∠BOD = ∠DOX = α

Therefore, ODOB = sin α

or, pOB = sin α

or, OB = psinα

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

xOA + yOB = 1

⇒ xpcosα + ypsinα = 1

⇒ xcosαp + ysinαp = 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 ∙ 1√2 + y ∙ 1√2 = 7

⇒ x√2 + 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

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