Question

The length of the perpendicular from the origin to a line is $7$ and the line makes an angle of ${150}^o$ with the positive direction of $y$ axis. The equation of the line is:

• A. $\sqrt{3}x+y=14$
• B. $\sqrt{3}x-y=13$
• C. $\sqrt{3}x+y+14=0$
• D. $\sqrt{3}x-y+13=0$

Answer 1

Answer: (A), (C)

The correct options are
A $\sqrt{3}x+y=14$
C $\sqrt{3}x+y+14=0$

Angle made with positive y-axis $={150}^o$

$\therefore$ Angle made with positive x-axis $={60}^o$

and ${180}^o-{60}^o={120}^o$

Let intercepts made by the line on x & y axis be a & b respectively,

Coordinates are $A(a,0)$ & $B(0,b)$

Slope $=\tan {120}^o$ & slope $=\tan {60}^o$

$\Rightarrow \frac{b-0}{0-a}=-\sqrt{3}$ $\Rightarrow \frac{b-0}{0-a}=\sqrt{3}$

$\Rightarrow b=a\sqrt{3}$ ……..$\left(1\right)$ $\Rightarrow b=-a\sqrt{3}$ …….$\left(2\right)$

Equation of line:

$y-0=\frac{b-0}{0-a}(x-a)$

$\Rightarrow y=-\sqrt{3}(x-a)$ & $y=\sqrt{3}(x-a)$

$\Rightarrow \sqrt{3}x+y-a\sqrt{3}=0$ & $\sqrt{3}x-y-a\sqrt{3}=0$

Now,

$d_{origin}=7$

$\Rightarrow \left|\frac{Ax+By+C}{\sqrt{A^2+B^2}}\right|=7$

$\left|\frac{\sqrt{3}.0\pm 1.0-a\sqrt{3}}{\sqrt{3+1}}\right|=7$

$\Rightarrow a\sqrt{3}=\pm 14$

$\therefore$ equation of lines are:

$\sqrt{3}x+y\pm 14=0$ & $\sqrt{3}x-y\pm 14=0$.