Question

Find the equation of the straight line passing through $(3,-2)$ and making an angle of ${60}^{\circ }$ with the positive direction of $y$-axis.

Answer 1

The required equation of the line is
$y-y_1=m(x-x_1)$

Since the line makes an angle ${60}^{\circ }$ with the positive direction of $y$-axis, it makes ${30}^{\circ }$ with the positive direction of $x$-axis.
$\therefore m=\tan {30}^{\circ }=\frac{1}{\sqrt{3}}$ [angle with $y$-axis]
A point on the line is $(x_1,y_1)=(3,-2)$
Therefore, the equation of the line is:
$y-y_1=m(x-x_1)⟹y-(-2)=\frac{1}{\sqrt{3}}(x-3)⟹y+2=\frac{x-3}{\sqrt{3}}⟹\sqrt{3}(y+2)=x-3⟹\sqrt{3}y+2\sqrt{3}=x-3⟹x-\sqrt{3}y-3-2\sqrt{3}=0⟹x-\sqrt{3}y-(3+2\sqrt{3})=0$