Coordinate Geometry - Straight Line

Find the equation of a straight line which passes through the point $\left(-1, 2\right)$ and is inclined at $60^\circ$ to the $X$ axis.


Inclination of the required line to the $X$ axis $= \theta = 60^\circ$

$\therefore \;$ Slope of the required line $= m = \tan \theta = \tan 60^\circ = \sqrt{3}$

The required line passes through the point $\left(x_1, y_1\right) = \left(-1, 2\right)$

Equation of the required line is of the form: $\;$ $y - y_1 = m \left(x - x_1\right)$

$\therefore \;$ The equation of the required line is

$y - 2 = \sqrt{3} \left(x + 1\right)$

i.e. $\;$ $\sqrt{3} x - y + \sqrt{3} + 2 = 0$