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,

\[\alpha = {60}^\circ\]

So, the equation of the line AB is

\[xcos\alpha + ysin\alpha = p \]

\[ \Rightarrow x\cos {60}^\circ + y\sin {60}^\circ = p\]

\[ \Rightarrow \frac{x}{2} + \frac{\sqrt{3}y}{2} = p\]

\[ \Rightarrow x + \sqrt{3}y = 2p . . . (1)\]

Now, in triangles OLA and OLB

\[\cos {60}^\circ = \frac{OL}{OA} \text { and } \cos {30}^\circ = \frac{OL}{OB}\]

\[ \Rightarrow \frac{1}{2} = \frac{p}{OA} \text { and } \frac{\sqrt{3}}{2} = \frac{p}{OB}\]

\[ \Rightarrow OA = 2p \text { and } OB = \frac{2p}{\sqrt{3}}\]

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

\[\therefore \frac{1}{2} \times OA \times OB = 96\sqrt{3}\]

\[ \Rightarrow \frac{1}{2} \times 2p \times \frac{2p}{\sqrt{3}} = 96\sqrt{3}\]

\[ \Rightarrow p^2 = {12}^2 \]

\[ \Rightarrow p = 12\]

Substituting the value of p in (1) 

\[x + \sqrt{3}y = 24\]

Hence, the equation of the line AB is

\[x + \sqrt{3}y = 24\]