$\begin {array} {1 1} (A)\;y + 2 = 0, \sqrt 3 x – y – 2 – 3 \sqrt 3 = 0 & \quad (B)\;x-2=0, \sqrt 3 x – y + 2 + 3 \sqrt 3 = 0 \\ (C)\;\sqrt 3 x – y – 2 – 3 \sqrt 3 = 0 & \quad (D)\;\text{None of these} \end {array}$

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- Equation of a line passing through a point $(x_1, y_1)$ and having slope $m$ is $ (y-y_1) = m (x-x_1)$

Step 1 :

The angle between the lines is given as $60^{\circ}$ and $ 0^{\circ}$.

Hence the slope 'm' is $ \tan 60^{\circ} = \sqrt 3 $ and $ \tan 0 = 0$

The line passing through the point (3,-2)

Hence the equation of the line is when $m =\sqrt 3 $

$ y-(-2)=\sqrt 3 (x-3)$

$ y+2=\sqrt 3x-3\sqrt 3$

$ \Rightarrow \sqrt 3x-y-2-3\sqrt 3 $

The equation of the line when $m = 0 $ is $y-(-2)=0$

$ \Rightarrow y + 2 = 0$

Hence the equation of the required lines are

$ y+2=0$ and $\sqrt 3 x-y-2-3\sqrt 3$

Hence 'A' is the correct option.

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