Question

Find the equation of the line having an inclination ${30}^{\circ }$ with the positive direction of $X$axis and cut off an intercept of $\frac{3}{2}$ on the positive side of $Y$ axis.

Answer 1

The intercept is cut off on the positive side of $Y$ axis

$\therefore c=\frac{3}{2}$

Slope$=m=\tan \theta$

$m=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$

Equation of the line is $y=mx+c$

$y=\frac{1}{\sqrt{3}}\times x+\frac{3}{2}$

$\therefore$ Equation of the line is $\frac{1}{\sqrt{3}}x-y+6=0$ or $x-\sqrt{3}y+6\sqrt{3}=0$