If the axes are rotated through an angle ${45}^0$ in the anti-clockwise direction, the coordinates of $(\sqrt{2},-\sqrt{2})$ in the new system are

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }={45}^0, \mathrm{and} \mathrm{P}(\mathrm{x},\mathrm{y})=(\sqrt{2},-\sqrt{2}) \\ \mathrm{let} (X,Y) \mathrm{be} \mathrm{new} \mathrm{coordinates} \mathrm{of} \mathrm{P} \mathrm{then} \\ \mathrm{X} =\mathrm{x} \cos \mathrm{\theta }+\mathrm{y} \sin \mathrm{\theta } \\ =\sqrt{2} \cos {45}^0-\sqrt{2} \sin 45 \\ =\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \\ =0 \\ \mathrm{and} Y=-\mathrm{x} \sin \mathrm{\theta }+\mathrm{y} \cos \mathrm{\theta } \\ =-\sqrt{2} \sin {45}^0-\sqrt{2} \cos {45}^0 \\ =-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)-\sqrt{2}\left(\frac{1}{\sqrt{2}}\right) \\ =-1-1 \\ =-2 \\ \therefore \mathrm{Required} \mathrm{point} = (0, -2) \end{gathered}$$