A straight line through the point (3,-2) is ndined at an angle $60^{\circ}$ to the line $\sqrt{3} x+y=1$. If it intersects the $X$ -axis, then its equation vill be

Given line, $$ \begin{array}{l} \sqrt{3} x+y=1 \\ y=-\sqrt{3} x+c \end{array} $$ We know that, $m=-\sqrt{3}$ $$ \begin{aligned} \text { that, } m &=-\sqrt{3} \quad[\because y=m x+c] \\ \tan \theta &=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right| \\ \tan 60^{\circ} &=\left|\frac{m_{1}-(-\sqrt{3})}{1+m_{1}(-\sqrt{3})}\right| \quad\left[\because \theta=60^{\circ}\right] \\ \sqrt{3}=&\left|\frac{m_{1}+\sqrt{3}}{1-\sqrt{3}} m_{1}\right| \\ \pm \sqrt{3}=& \frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} \end{aligned} $$ \begin{array}{l|l} \hline taking + sign & taking - sign \\ \hline+\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} & -\sqrt{3}=\frac{m_{1}+\sqrt{3}}{1-\sqrt{3} m_{1}} \\ \sqrt{3}\left(1-\sqrt{3} m_{1}\right)=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}\left(1-\sqrt{3} m_{1}\right) \\ & =m_{1}+\sqrt{3} \\ \Rightarrow \sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} & \Rightarrow-\sqrt{3}-3 m_{1}=m_{1}+\sqrt{3} \\ \Rightarrow & \Rightarrow 3 m_{1}-m_{1}=2 \sqrt{3} \\ \Rightarrow 4m_{1}=0 & \Rightarrow \quad 2 m_{1}=2 \sqrt{3} \\ \Rightarrow m_{1}=0 & \Rightarrow \quad m_{1}=\sqrt{3} \end{array} given point (3,-2) $y=m x$ + c $$ \begin{aligned} -2 &=3 \sqrt{3}+c \\ c &=-2-3 \sqrt{3} \end{aligned} $$ Hence, required equation $$ \therefore \quad y=m x+c $$ $$ \begin{array}{l} y=\sqrt{3} x-2-3 \sqrt{3} \\ y-\sqrt{3} x+2+3 \sqrt{3}=0 \\ y-x \sqrt{3}+2+3 \sqrt{3}=0 \end{array} $$