Find the values of θ and p, if the equation x cos θ + y sin θ = p is the normal form of the line $$\sqrt{3}x+y-2=0$$

Asked by Abhisek | 1 year ago |  117

##### Solution :-

The equation of the given line is $$\sqrt{3}x+y+2=0$$

This equation can be reduced as,

$$-\sqrt{3}x-y=2$$

Divide both sides by

$$\sqrt{(-\sqrt{3})^2+(-1)^2}=2$$,

We get,

$$\dfrac{-\sqrt{3}}{2}x-\dfrac{1}{2}y=\dfrac{2}{2}$$

$$\dfrac{-\sqrt{3}}{2}x-\dfrac{1}{2}y=1$$ ...........(1)

Compare equation (1) to xcos θ+ ysin θ= p , we get

cos θ = $$\dfrac{-\sqrt{3}}{2}$$, sin θ = $$- \dfrac{1}{2}$$, and p=1

Therefore, the respective values of θ and p are

$$\dfrac{7\pi}{6}$$ and 1.

Answered by Pragya Singh | 1 year ago

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