First slide

Cartesian plane

Question

A line forms a triangle of area $54 \sqrt{3}$ sq-units with the coordinate axes. If the perpendicular drawn from the origin to the line makes an angle of $60 °$ with the $x -$ axis, then the equation the line is

Moderate

A

$\sqrt{3} x + y = 18$
B

$x + \sqrt{3} y = 9$
C

$x + \sqrt{3} y = 18$
D

none of these

Solution

Let $p$ be the length of the perpendicular drawn from the origin to the required line. The perpendicular makes an angle of $60 °$ with the $x -$ axis. Putting $a = 60 °$ in $x \cos α + y \sin α = p$ the equation of the line is $x \cos 60^{\circ} + y \sin 60^{\circ} = p$
or, $x + \sqrt{3} y = 2 p$ .
This cuts the coordinate axes at $A (2 p, 0)$ and $B (0, \frac{2 p}{\sqrt{3}})$
It is given that
Area of $△ O A B = 54 \sqrt{3}$ sq. units.
$\begin{array}{l} \Rightarrow \frac{1}{2} (O A) (O B) = 54 \sqrt{3} \\ \Rightarrow \frac{1}{2} \times 2 p \times \frac{2 p}{\sqrt{3}} = 54 \sqrt{3} \Rightarrow p^{2} = 81 \Rightarrow p = 9 \end{array}$
Hence, the equation of the line is $x + \sqrt{3} y = 18$

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