Derive the equation of the line in normal form in terms of coordinate.
Chapter → Straight lines
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Answers
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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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