Question

The equation of the lines on which the perpendicular from the origin make ${30}^{\circ }$ angle with $x-$axis and which form a triangle of area $\frac{50}{\sqrt{3}}$sq.units with axes are

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

Answer 1

Answer: (A), (B)

The correct options are
A $\sqrt{3}x+y+10=0$
B $\sqrt{3}x+y-10=0$

Let the perpendicular distance of the line from the origin be x
Let the x-intercept made by the line be $\lambda$ and y-intercept be $\omega$

The general equation of line thus would be
$y=tan({180}^0-{60}^0)x+\omega$

$y=-\sqrt{3}+\omega$ ...(i) (since the angle made by the perpendicular is ${30}^0$, therefore, the angle made by the line with x-axis would be ${90}^0-{30}^0={60}^0$)

By geometry $\omega sin{30}^0=p$ and $\lambda cos{30}^0=p$

$\omega =2p$ and $\lambda =\frac{2p}{\sqrt{3}}$

Now the area the line makes with the coordinate axes would be
$\frac{(x-intercept)(y-intercept)}{2}$

$=\frac{\lambda \left(\omega \right)}{2}$

$=\frac{4p^2}{2\sqrt{3}}$

$=\frac{50}{\sqrt{3}}$ ...(given)

Hence $p=\pm 5$

Therefore $\omega$ will be $\frac{p}{sin{30}^0}$

$=\pm 10$

Substituting in the original equation we get

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

Hence we get a pair of lines