Question

A line cuts off an intercept of $4$ units on negative direction of the y-axis and makes an angle of ${120}^o$ with the positive direction of the x-axis. The equation of the line is

• A. $\sqrt{3}x-y-4=0$
• B. $\sqrt{3}x+y-4=0$
• C. $\sqrt{3}x+y+4=0$
• D. none of these

Answer 1

Answer: (C)

$\sqrt{3}x+y+4=0$

Here, $m=\tan {120}^o=\tan ({180}^o-{60}^o)=-\tan {60}^o=-\sqrt{3};c=-4$

the required equation is $y=-\sqrt{3}-4\Rightarrow \sqrt{3}x+y+4=0$