Question

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

• A. $x+\sqrt{3}y\pm 10=0$
• B. $\sqrt{3}x+y\pm 10=0$
• C. $x\pm \sqrt{3}y-10=0$
• D. None of the above

Answer 1

The correct option is B $\sqrt{3}x+y\pm 10=0$

Let $p$ be the length of perpendicular from the origin on the given line.

Then its equation in normal form is
$x\cos {30}^{\circ }+y\sin {30}^{\circ }=p$ or $\sqrt{3}x+y=2p$

This meets with the coordinate axes at
$A(\frac{2p}{\sqrt{3}},0)$ and $B(0,2p)$.

$\therefore$ Area of $△OAB=\frac{1}{2}\left(\frac{2p}{\sqrt{3}}\right)2p=\frac{2p^2}{\sqrt{3}}$

By hypothesis $\frac{2p^2}{\sqrt{3}}=\frac{50}{\sqrt{3}}\Rightarrow p=\pm 5$

Hence, the lines are

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