A straight line L through the point ( 3, - 2 ) is inclined at an...

Let $m$ be the slope of the line,

then $tan {60}^{o} = \left|\dfrac{m + \sqrt{3}}{1 - m\sqrt{3}}\right|<br /><br />$⇒ \sqrt{3} = \left|\dfrac{m + \sqrt{3}}{1 - \sqrt{3}m}\right|$<br /><br />$⇒ \pm \sqrt{3} = \dfrac{m + \sqrt{3}}{1 - \sqrt{3}m}$</p><p>$∴ \sqrt{3} - 3m = m + \sqrt{3}$or$- \sqrt{3} + 3m = m + \sqrt{3}$<br /><br />4m = 0 or$2m = 2\sqrt{3}$<br /><br />m = 0 or$m = \sqrt{3}$<br /><br />Since L intersects$X$-axis we take$m = \sqrt{3}$<br /><br />and get$y = \sqrt{3}x + c$<br /><br />Since it passes through \left(3, - 2\right)$, therefore line will become $y - \sqrt{3}x + 2 + 3\sqrt{3} = 0$