Question

Find the equation of a straight line on which the perpendicular from the origin makes an angle of 30° with x-axis and which forms a triangle of area $50/\sqrt{3}$ with the axes.

Answer 1

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



Here, $\mathrm{\alpha }={30}^{\circ }$

So, the equation of the line AB is

$$\begin{gathered} x\mathrm{cos\alpha }+y\mathrm{sin\alpha }=p \\ \Rightarrow x\cos {30}^{\circ }+y\sin {30}^{\circ }=p \\ \Rightarrow \frac{\sqrt{3}x}{2}+\frac{y}{2}=p \\ \Rightarrow \sqrt{3}x+y=2p...\left(1\right) \end{gathered}$$

Now, in triangles OLA and OLB

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

It is given that the area of triangle OAB is $50/\sqrt{3}$

$$\begin{gathered} \therefore \frac{1}{2}\times OA\times OB=\frac{50}{\sqrt{3}} \\ \Rightarrow \frac{1}{2}\times \frac{2p}{\sqrt{3}}\times 2p=\frac{50}{\sqrt{3}} \\ \Rightarrow p^2=25 \\ \Rightarrow p=5 \end{gathered}$$

Substituting the value of p in (1):

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

Hence, the equation of the line AB is $x+\sqrt{3}y=10$.