Question

The length of perpendicular from the origin to a line is $5$ units and the line makes an angle ${120}^{\circ }$ with the positive direction of x-axis. The equation of the line is.

• A. $x+\sqrt{3}y=5$
• B. $\sqrt{3}x+y=10$
• C. $\sqrt{3}x-y=10$
• D. None of the above

Answer 1

The correct option is B $\sqrt{3}x+y=10$

Given that $OC=5$

Angle made by line $AB$ with positive x axis is ${120}^o$

So, $\angle CAO=180-120={60}^o$

In triangle $OCA$ sum of angles $=180$

$\angle OCA+\angle CAO+\angle AOC=180⟹90+60+\angle AOC=180⟹\angle AOC={30}^o$

Slope of the line $OC=\tan {30}^o=\frac{1}{\sqrt{3}}$

Length of $OC=5$

Hence the point $C$ which is in first quadrant will be

$(0+5\cos {30}^o,0+5\sin {30}^o)⟹C=(\frac{5\sqrt{3}}{2},\frac{5}{2})$

The line having slope $-\sqrt{3}$ and passing through $C=(\frac{5\sqrt{3}}{2},\frac{5}{2})$ shall become

$y-\frac{5}{2}=-\sqrt{3}(x-\frac{5\sqrt{3}}{2})⟹\sqrt{3}x+y=10$