Question

Find the equation of the straight line which makes a triangle of area $96\sqrt{3}$ with the axes and perpendicular from the origin to it makes an angle of 30° with Y-axis.

Answer 1

Let AB be the given line and OL = p be the perpendicular drawn from the origin on the line.


Here, $\mathrm{\alpha }={60}^{\circ }$

So, the equation of the line AB is

$$\begin{gathered} x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p \\ \Rightarrow x\cos {60}^{\circ }+y\sin {60}^{\circ }=p \\ \Rightarrow \frac{x}{2}+\frac{\sqrt{3}y}{2}=p \\ \Rightarrow x+\sqrt{3}y=2p...\left(1\right) \end{gathered}$$

Now, in triangles OLA and OLB

$$\begin{gathered} \cos {60}^{\circ }=\frac{OL}{OA}\mathrm{and}\cos {30}^{\circ }=\frac{OL}{OB} \\ \Rightarrow \frac{1}{2}=\frac{p}{OA}\mathrm{and}\frac{\sqrt{3}}{2}=\frac{p}{OB} \\ \Rightarrow OA=2p\mathrm{and}OB=\frac{2p}{\sqrt{3}} \end{gathered}$$

It is given that the area of triangle OAB is $96\sqrt{3}$

$$\begin{gathered} \therefore \frac{1}{2}\times OA\times OB=96\sqrt{3} \\ \Rightarrow \frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=96\sqrt{3} \\ \Rightarrow p^2={12}^2 \\ \Rightarrow p=12 \end{gathered}$$

Substituting the value of p in (1)

$x+\sqrt{3}y=24$

Hence, the equation of the line AB is $x+\sqrt{3}y=24$