Question

The equation of the lines on which the perpendicular from the origin make ${30}^o$ angle with xxaxis and which form a triangle of area $\frac{50}{\sqrt{3}}$ with axes are

• A. $x+\sqrt{3}y\pm 10=0$
• B. $\sqrt{3}x+y\pm 10=0$
• C. $x\pm \sqrt{3}y-10=0$
• D. $Noneofthese$

Answer 1

Answer: (B)

$\sqrt{3}x+y\pm 10=0$

$AB$ is the given line and $OL=p$ is the perpendicular drawn from the origin to the line $OL$ makes an angle of ${30}^{\circ }$ with $OX$

$\therefore \alpha ={30}^{\circ }$

equation of line $AB$ is given by

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

Now, In $\Delta OLA$

$\begin{array}{c}\cos {30}^{\circ }=\frac{OL}{OA}\\ \Rightarrow OA=\frac{2p}{\sqrt{3}}\end{array}$

Now, In $\Delta OLB$

$\begin{array}{c}\cos {60}^{\circ }=\frac{OL}{OB}\\ \Rightarrow OB=2p\end{array}$

Given that,

$\begin{array}{c}Area=\frac{\sqrt{50}}{3}\\ \Rightarrow \frac{1}{2}\times \frac{2p}{\sqrt{3}}\times 2p=\frac{50}{\sqrt{3}}\\ \Rightarrow p^2=25\\ \therefore p=\pm 5\end{array}$

put the value in equation $\left(1\right)$

hence the equation of line $AB$ is given by $\sqrt{3x}+y=\pm 10$