Find the values of and p, if the equation x + y = p is the normal form of the line 3 x + y + 2 = 0

Here 3 x + y + 2 = 0 3 x + y = - 2 - 3 x - y = 2 Dividing both sides by ( - 3 ) ^2 + ( - 1) ^2 = 2, we have - 3 2 x - 1 2 y = 1 Put = - 3 2 and = - 1 2 lies in 3rd quadrant = - 3 2 = - 30^ = cos (180° + 30° ) = 210^ Equation of line in normal form is x 7 6 + y 7 6 = 1 Comparing it with x cos + y sin = p, we have = 7 6 and p = 1

Find the values of and p, if the equation x + y = p is the normal…

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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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Here $\sqrt 3 x + y + 2 = 0$
$ \Rightarrow \sqrt 3 x + y = - 2 \Rightarrow - \sqrt 3 x - y = 2$
Dividing both sides by $\sqrt {{{( - \sqrt 3 )}^2} + {{( - 1)}^2}} = 2$, we have
$\frac{{ - \sqrt 3 }}{2}x - \frac{1}{2}y = 1$
Put $\cos \alpha = \frac{{ - \sqrt 3 }}{2}$ and $\sin \alpha= \frac{{ - 1}}{2}$
$ \Rightarrow \alpha $ lies in 3rd quadrant
$\therefore \cos \alpha = \frac{{ - \sqrt 3 }}{2} = - \cos 30^\circ $ = cos (180° + 30° )
$ \Rightarrow \alpha = 210^\circ $
$\therefore $ Equation of line in normal form is
$x\cos \frac{{7\pi }}{6} + y\sin \frac{{7\pi }}{6} = 1$
Comparing it with x cos $\alpha + y\ sin \alpha = p$, we have
$\alpha = \frac{{7\pi }}{6}$ and p = 1

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