Question

Equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of ${120}^{\circ }$ with $x-$axis is

• A. $x\sqrt{3}+y+8=0$
• B. $x+\sqrt{3}y=8$
• C. $x\sqrt{3}-y=8$
• D. $x-\sqrt{3}y+8=0$

Answer 1

The correct option is A $x\sqrt{3}+y+8=0$

Using the formula of perpendicular distance of any point $(x_0,y_0)$ from line $ax+by+c=0$ is $\frac{|a{x}_0+b{y}_0+c|}{\sqrt{a^2+b^2}}$

Length of perpendicular from origin on line

$ax+by+c=0$ is $\left|\frac{a\left(0\right)+b\left(0\right)+c}{\sqrt{a^2+b^2}}\right|=\left|\frac{c}{\sqrt{a^2+b^2}}\right|=4units$ $\Rightarrow \frac{c^2}{a^2+b^2}=16....\left(i\right)$

Line $ax+by+c=0$ has slope $m=\tan {120}^2=-\sqrt{3}=-\frac{a}{b}$ $\Rightarrow a^2=3b^2........\left(ii\right)$

putting (ii) in (i)we get $c^2=16\left(4b^2\right)$ $⟹c=\pm 8b$

Putting values of a,c in terms of b in the equation of line we get,

$-\sqrt{3}bx+by\pm 8b=0⟹-\sqrt{3}x+y\pm 8=0$