Question

The equation 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 these

Answer 1

Answer: (B)

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

Answer 2

Let p be the length of the 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 $\sqrt { 3 } x + y = 2 p$

This meets the coordinate axes at $A \left( \frac { 2 p } { \sqrt { 3 } } , 0 \right)$ and $B ( 0,2 p )$. ∴ Area of

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

Hence the lines are $\sqrt { 3 } x + y \pm 10 = 0$.