Find the equation of the line which is at a p | vedclass

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Question

(A) The normal form of the equation of a line is given by $x \cos \omega + y \sin \omega = p$,where $p$ is the perpendicular distance from the origin

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$\sqrt{3}x + y = 10$

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(A) The normal form of the equation of a line is given by $x \cos \omega + y \sin \omega = p$,where $p$ is the perpendicular distance from the origin and $\omega$ is the angle made by the perpendicular with the positive $x-$ axis.Given $p = 5$ and $\omega = 30^{\circ}$.Substituting these values into the equation:$x \cos 30^{\circ} + y \sin 30^{\circ} = 5$Since $\cos 30^{\circ} = \frac{\sqrt{3}}{2}$ and $\sin 30^{\circ} = \frac{1}{2}$,we have:$x \left( \frac{\sqrt{3}}{2} \right) + y \left( \frac{1}{2} \right) = 5$Multiplying both sides by $2$,we get:$\sqrt{3}x + y = 10$

Find the equation of the line which is at a perpendicular distance of $5$ units from the origin and the angle made by the perpendicular with the positive $x-$ axis is $30^{\circ}$.

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