Find the value of θ and p if the equation x cos θ + y sin θ = p is

1.	Find the value of θ and p if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.
Answer» Given: the normal form of a line is x cos θ + y sin θ = p …..… (1) To find: P and θ. Explanation: Let us try to write down the equation √3 + y + 2 = 0 in its normal form. Now √3 + y + 2 = 0 ⇒ √3 + y = – 2 Dividing both sides by 2, ⇒ – √3/2 – y/2 = 1 ⇒ \(\Big(-\frac{\sqrt{3}}{2}\Big)x+\Big(-\frac{1}{2}\Big)y\) = 1 ......(2) Comparing equations (1) and (2) we get, cosθ = \(-\frac{\sqrt{3}}{2}\) and p = 1 ⇒ θ = 210° = 7π/6 and p = 1 Hence, θ = 210° = 7π/6 and p = 1

Find the value of θ and p if the equation x cos θ + y sin θ = p is the normal form of the line √3x + y + 2 = 0.

Answer»

Given: the normal form of a line is x cos θ + y sin θ = p …..… (1)

To find: P and θ.

Explanation:

Let us try to write down the equation √3 + y + 2 = 0 in its normal form.

Now √3 + y + 2 = 0

⇒ √3 + y = – 2

Dividing both sides by 2,

⇒ – √3/2 – y/2 = 1

⇒ \(\Big(-\frac{\sqrt{3}}{2}\Big)x+\Big(-\frac{1}{2}\Big)y\) = 1 ......(2)

Comparing equations (1) and (2) we get,

cosθ = \(-\frac{\sqrt{3}}{2}\) and p = 1

⇒ θ = 210° = 7π/6 and p = 1

Hence, θ = 210° = 7π/6 and p = 1