Question

A line forms a triangle of area $54\sqrt{3}$ square units with the coordinate axes. Find the equation of the line if the perpendicular drawn from the origin to the line makes an angle of $60º$ with the$x-$axis

Answer 1

Step 1: Write the given values.

angle made with $x-$axis $\alpha =60°$

Area of triangle made by the line with the coordinate axes$=54\surd 3sq.units$

Step 2: Form the equation of the line with the given data.

the equation of the line if the perpendicular drawn from the origin to the line makes an angle of $\alpha$ with the$x-$axis is

$x\cos \alpha +y\sin \alpha =p$ where $p$ is the perpendicular distance from the origin

Therefore,the equation of the line is

$x\cos 60°+y\sin 60°=p$

$\Rightarrow x+\surd 3y=2p$

$\Rightarrow x+\surd 3y=2p$

Step 3: Find the value of $p$

The area of the triangle formed by the line with the coordinate axes $=54\surd 3sq.units$

$$\begin{gathered} \Rightarrow \frac{1}{2}\times \frac{{\left(2p\right)}^2}{1\times \sqrt{3}}=54\sqrt{3} \\ \Rightarrow 2p^2=54\times 3 \\ \Rightarrow p^2=81 \\ \Rightarrow p=\sqrt{81}=9 \end{gathered}$$

Hence, the equation of the line is $x+\surd 3y=18$