If the lines represented by the equation $3{\mathrm{y}}^2-{\mathrm{x}}^2+2\sqrt{3}\mathrm{x}-3=0$ are rotated about the point $(\sqrt{3},0)$ through an angle of ${15}^0,$ one in clockwise direction and the other in anticlockwise direction, so that they become perpendicular, then the equation of the pair of lines in the new position is

The given equation of the pair of straight lines can be rewritten as $(\sqrt{3}\mathrm{y}-\mathrm{x}+\sqrt{3})(\sqrt{3}\mathrm{y}+\mathrm{x}-\sqrt{3})=0$

Their separate equations are $\sqrt{3}\mathrm{y}-\mathrm{x}+\sqrt{3}=0$ and $\sqrt{3}\mathrm{y}+\mathrm{x}-\sqrt{3}=0$

or $\mathrm{y}=\frac{1}{\sqrt{3}}\mathrm{x}-1$ and $\mathrm{y}=-\frac{1}{\sqrt{3}}\mathrm{x}+-1$ 

or $\mathrm{y}=\left(\tan {30}^0\right)\mathrm{x}-1$ and $\mathrm{y}=\left(\tan {150}^0\right)\mathrm{x}+1$

After rotation through an angle of ${15}^0$ as given in the question, the lines are $(\mathrm{y}-0)=\tan {45}^0(\mathrm{x}-\sqrt{3})$ and $(\mathrm{y}-0)=\tan {135}^0(\mathrm{x}-\sqrt{3})$

or $\mathrm{y}=\mathrm{x}-\sqrt{3}$ $\mathrm{y}=-\mathrm{x}+\sqrt{3}$ Their combined equation is $(\mathrm{y}-\mathrm{x}+\sqrt{3})(\mathrm{y}+\mathrm{x}-\sqrt{3})=0$

or ${\mathrm{y}}^2-{\mathrm{x}}^2+2\sqrt{3}\mathrm{x}-3=0$