Question

Find the equation of the lines on which the perpendiculars from the origin make ${30}^0$ angles with x-axis and which form the triangle of area $\frac{50}{\sqrt{3}}$ with axes. Enter 1 if answer is $\sqrt{3}x+y=\pm 10$ else enter 0.

Answer 1

Given area of $△OAB=\frac{50}{\sqrt{3}}$

$OA\perp$ line $m$. and $\angle AOB={30}^o$

Let point of intersection of line $m$ with x-axis is $(a,0)$

Then $OB=a$

From $△OAB$ $OA=\frac{\sqrt{3}a}{2}$ & $AB=\frac{a}{2}$.

Now area of $△OAB=\frac{1}{2}\times OA\times AB$

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

$a^2=\frac{8\times 50}{3}=\frac{400}{3}$

$a=\frac{20}{\sqrt{30}}$

Now, slope of line $m=\tan {160}^o=-\sqrt{3}$

and point $B$ is $(\frac{20}{\sqrt{3}},o)$

Hence equation of line is

$(y-0)=m(x-a)$

$y-0=-\sqrt{3}\left(\right(x-\frac{20}{\sqrt{3}})$

$y+\sqrt{3}x=20$

Hence equation of line is $y+\sqrt{3}x=20$