Find the values of theta and p,if the equatio | vedclass

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(A) The equation of the given line is $\sqrt{3} x + y + 2 = 0$. This equation can be rewritten as $-\sqrt{3} x - y = 2$. Dividing both sides by $\sqrt

Vedclass · Accepted Answer

$	heta = \frac{7 \pi}{6}, p = 1$

Vedclass · Answer

(A) The equation of the given line is $\sqrt{3} x + y + 2 = 0$. This equation can be rewritten as $-\sqrt{3} x - y = 2$. Dividing both sides by $\sqrt{(-\sqrt{3})^2 + (-1)^2} = \sqrt{3 + 1} = 2$,we get: $-\frac{\sqrt{3}}{2} x - \frac{1}{2} y = 1$. Comparing this with the normal form $x \cos 	heta + y \sin 	heta = p$,we have: $\cos 	heta = -\frac{\sqrt{3}}{2}$,$\sin 	heta = -\frac{1}{2}$,and $p = 1$. Since both $\sin 	heta$ and $\cos 	heta$ are negative,$	heta$ lies in the third quadrant. $	heta = \pi + \frac{\pi}{6} = \frac{7 \pi}{6}$. Thus,the values are $	heta = \frac{7 \pi}{6}$ and $p = 1$.

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

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