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 ${30}^{\circ }$ with y-axis.

Answer 1

$\text{Perpendicular from origin makes an angle of}{30}^{\circ }\text{with y-axis, thus making}{60}^{\circ }\text{with x-axis}$

$\text{Area of triangle is}=96\sqrt{3}$

$In\Delta OLAand\Delta OLB$

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

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

$OA=2pandOB=\frac{p}{\sqrt{3}}$

$\text{Area of triangle}=\frac{1}{2}\times OA\times OB$

$\therefore \frac{1}{2}\times 2p\times \frac{2p}{\sqrt{3}}=96\sqrt{3}$

$=\frac{2p^2}{\sqrt{3}}=96\sqrt{3}$

$p^2=\frac{96\sqrt{3}\times \sqrt{3}}{2}=48\times 3$

$=2\times 2\times 2\times 2\times 3\times 3$

$p=12$

$xcos\alpha +ysin\alpha =p$

$xcos{60}^{\circ }+ysin{60}^{\circ }=12$

$x\times \frac{1}{2}+y\frac{\sqrt{3}}{2}=12$

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