Question

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

Answer 1

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

$\text{It is given that,}$

$\text{area of triangle}=\frac{50}{\sqrt{3}}$

$\text{In triangle OLA and OLB}$

$cos{30}^{\circ }=\frac{OL}{OA}andcos{60}^{\circ }=\frac{OL}{OB}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{p}{OB}and\frac{1}{2}=\frac{p}{OB}$

$\Rightarrow OA=\frac{2p}{\sqrt{3}}andOB=2p$

$\text{are of triangle}=\frac{1}{2}{r}^{2}sin\theta =\frac{50}{\sqrt{3}}$

$sin30=\frac{1}{2}$

$\frac{1}{2}\times OA\times OB=\frac{50}{\sqrt{3}}$

$\frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=\frac{50}{\sqrt{3}}$

$p^2=\frac{50}{\sqrt{3}}\times \frac{\sqrt{3}}{2}=25$

$p\pm 5$

$\text{Now, equation of the line is x}cos\alpha +y$

$sin\alpha =p$

$xcos\alpha +ysin\alpha =\pm 5$

$xcos{30}^{\circ }+ysin{30}^{\circ }=\pm 5$

$\text{By substituting the values,}$

$x\frac{\sqrt{3}}{2}+\frac{y}{2}=\pm 5$

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

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