Find the equation of the line such that the p | vedclass

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(B) Let the length of the perpendicular from the origin to the line be $p$. The normal form of the equation is $x \cos 30^{\circ} + y \sin 30^{\circ}

Vedclass · Accepted Answer

$\sqrt{3}x + y \pm 10 = 0$

Vedclass · Answer

(B) Let the length of the perpendicular from the origin to the line be $p$. The normal form of the equation is $x \cos 30^{\circ} + y \sin 30^{\circ} = p$,which simplifies to $\frac{\sqrt{3}}{2}x + \frac{1}{2}y = p$,or $\sqrt{3}x + y = 2p$.This line intersects the axes at points $A\left(\frac{2p}{\sqrt{3}}, 0ight)$ and $B(0, 2p)$.The area of $\Delta OAB$ is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes \left|\frac{2p}{\sqrt{3}}ight| 	imes |2p| = \frac{2p^2}{\sqrt{3}}$.Given the area is $\frac{50}{\sqrt{3}}$,we have $\frac{2p^2}{\sqrt{3}} = \frac{50}{\sqrt{3}}$,which implies $p^2 = 25$,so $p = \pm 5$.Substituting $p = \pm 5$ into $\sqrt{3}x + y = 2p$,we get $\sqrt{3}x + y = \pm 10$,or $\sqrt{3}x + y \mp 10 = 0$.

Find the equation of the line such that the perpendicular drawn from the origin to the line makes an angle of $30^{\circ}$ with the $x$-axis and the line forms a triangle of area $\frac{50}{\sqrt{3}}$ with the axes.

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