Various forms of equation of straight lines - I

Normal or perpendicular form of a line:

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 cosα + y sinα = P.

Example:

A line forms a triangle of area 54√3 sq units with the coordinate axes. Find the equation of the line if the perpendicular drawn from the origin to the line makes an angle of 60⁰ with the x – axis.

Solution:

Let p be the length of the perpendicular drawn from the origin to the required line. The perpendicular makes an angle of 60⁰ with the x-axis. So, the equation of the line is x cos60⁰ + y sin 60⁰ = p

Or, x + √3y = 2p.

This cuts the coordinate axes at A (2p, 0) and B (0, 2p/√3).

It is given that

Area of ∆OAB = 54√3 Squints

⇒ ½ (OA) (OB) = 54√3

⇒ ½ x 2p x 2p/√3 = 54√3

⇒ p² = 81

⇒ p = 9

Hence the equation of the line is x + √3y = 18.

Reduction of the general equation of a line to the normal form:

The general equation of a line is Ax + By + C = 0 and its normal form be x cosα + y sinα = P.

To reduce the general equation of a line to normal form, we first shift the constant term on the RHS and make it positive if it is not so and then divide both sides by √ (coeff. of x)² + (coeff. of y)²

i.e., [A/ √ (A² + B²)] x + [B/ √ (A² + B²)] y = [-C/ √ (A² + B²)].

Distance form of a line:

The equation of the straight line passing through (x₁, y₁) and making an angle θ with the positive direction of x-axis is (x - x₁)/ cosθ = (y - y₁) / sinθ = r

Where r is the distance of a point (x, y) on the line from the point (x₁, y₁)

The coordinates of any point on the line at a distance  from the given point (x₁, y₁) are (x₁ ± r cosθ, y₁ ± r sinθ). If the point is on the right side of (x₁, y₁), then r is positive and if the point is on the left side of(x₁, y₁), then r is negative.

Example:

A straight line is drawn through the point P (2, 3) and is inclined at an angle of 30⁰ with the x-axis. 

Find the coordinates of two points on it at a distance 4 from P on either side of P?

Solution:

Here, (x₁, y₁) = (2, 3) and θ = 30⁰. So, the equation of line is (x – 2) / cos 30⁰ = (y - 3)/ sin 30⁰

Or, (x - 2)/ [√3/2] = (y - 3)/ [½]  

Or, x-2 = √3(y – 3)

Or, x - √3y = 2 – 3√3

Points on the line at a distance 4 from P (2, 3) are given by (x₁ ± r cosθ, y₁ ± r sinθ)

Or, (2 ± 4 cos30⁰, 2 ± 4 sin30⁰)

Or, (2 ± 2√3, 2 ± 2)

Or, (2 + 2√3, 4) and (2 - 2√3, 0)