Question

Find the equation of a line which makes an angle of $$60^{\circ}$$ with the positive direction of $$x$$-axis and passes through the point $$\mathrm{P}(0, -3)$$.

Answer 1

The slope of the line is $$m = \tan 60^{\circ} = \sqrt{3}$$

Since the line passes through $$\mathrm{P}(0, -3)$$, this point lies on the $$y$$-axis, so the $$y$$-intercept is $$c = -3$$

Using the slope-intercept form $$y = mx + c$$:

$$y = \sqrt{3}x - 3$$

$$\sqrt{3}x - y - 3 = 0$$

Thus, the equation of the line is $$\sqrt{3}x - y - 3 = 0$$.