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(B) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the or

Vedclass · Accepted Answer

$\sqrt{3}x + y = 8$

Vedclass · Answer

(B) The normal form of the equation of a line is given by $x \cos \alpha + y \sin \alpha = p$,where $p$ is the length of the perpendicular from the origin and $\alpha$ is the angle the perpendicular makes with the positive $x$-axis.Given $p = 4$ and $\alpha = 30^{\circ}$.Substituting these values into the formula:$x \cos 30^{\circ} + y \sin 30^{\circ} = 4$$x \left( \frac{\sqrt{3}}{2} \right) + y \left( \frac{1}{2} \right) = 4$Multiplying the entire equation by $2$:$\sqrt{3}x + y = 8$.

The length of the perpendicular from the origin to a line is $4$ units,and the perpendicular makes an angle of $30^{\circ}$ with the positive direction of the $x$-axis. Find the equation of the line.

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