The equation of the lines on which the perpen | vedclass

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

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

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(B) Let $p$ be the length of the perpendicular from the origin to the line. The equation of the line in normal form 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 coordinate axes at points $A\left(\frac{2p}{\sqrt{3}}, 0ight)$ and $B(0, 2p)$.The area of the triangle $\Delta OAB$ formed with the axes is given by $	ext{Area} = \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 that 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 the equation $\sqrt{3}x + y = 2p$,we get $\sqrt{3}x + y = \pm 10$,or $\sqrt{3}x + y \mp 10 = 0$. Thus,the lines are $\sqrt{3}x + y \pm 10 = 0$.

The equation of the lines on which the perpendiculars from the origin make a $30^\circ$ angle with the $x$-axis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with the axes,are

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