Given: (x₁,y₁) = A ( – 2, – 7)

To find:

Equation of required line.

Explanation:

So, the equation of the line is

Formula Used:

Let the required line intersect the lines 4x + 3y = 3 and 4x + 3y = 12 at P₁ and P₂.

Let AP_{1 =} r₁ and AP₂ = r₂

Then, the coordinates of P₁ and P₂ are given by and respectively. Thus, the coordinates of P₁ and P₂ are ( – 2 + r₁cos θ, – 7 + r₁sin θ) and ( – 2 + r₂cos θ, – 7 + r₂sin θ), respectively.

Clearly, the points P₁ and P₂ lie on the lines 4x + 3y = 3 and 4x + 3y = 12

4( – 2 + r₁cos θ) + 3( – 7 + r₁sin θ) = 3 and 4( – 2 + r₂cos θ) + 3( – 7 + r₂sin θ) = 12,

and

Here AP₂ – AP₁ = 3

⇒ 3 = 4 cos θ + 3 sin θ

⇒ 3(1 – sin θ) = 4cos θ

⇒ 9(1 + sin² θ – 2sin θ) = 16 cos² θ = 16(1 – sin² θ)

⇒ 25 sin² θ – 18 sin θ – 7 = 0

⇒ (sin θ – 1)(25 sin θ + 7 ) = 0

⇒ Sin θ = 1, sin θ = – 7/25

⇒ Cos θ = 0, cos θ = 24/25

Thus, the equation of the required line is

x + 2 = 0 or

⇒ x + 2 = 0 or 7x + 24 y + 182 = 0

Hence, the equation of line is x + 2 = 0 or 7x + 24 y + 182 = 0