If the origin of a coordinate system is shifted to $(-\sqrt{2},\sqrt{2})$ and then the coordinate system is rotated anticlockwise through an angle ${45}^0$,
the point P(1, –1) in the original system has new coordinates

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{new} \mathrm{coordinates} \mathrm{of} \mathrm{the} \mathrm{point} \mathrm{be} (X, Y) \\ \mathrm{since} \mathrm{origin} \mathrm{is} \mathrm{shifted} \mathrm{to} (-\sqrt{2}, \sqrt{2}) \mathrm{and} \mathrm{rotates} \\ \mathrm{anti}-\mathrm{clock} \mathrm{wise} \mathrm{through} \mathrm{an} \mathrm{angle} {45}^{0.} \\ \mathrm{So}, (\mathrm{h}, \mathrm{k})=(-\sqrt{2}, \sqrt{2}), \mathrm{\theta }=\frac{\mathrm{\pi }}{4} \mathrm{and} (\mathrm{x}, \mathrm{y})=(1, -1) \\ \therefore X=(\mathrm{x}-\mathrm{h})\mathrm{cos\theta }+(\mathrm{y}-\mathrm{k})\mathrm{sin\theta } \\ =(1+\sqrt{2 })\cos {45}^0+(-1-\sqrt{2})\sin {45}^0 \\ = (1+\sqrt{2})\frac{1}{\sqrt{2}}+(-1-\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{1}{\sqrt{2}} (1+\sqrt{2}-1-\sqrt{2}) \\ =0 \\ X=0 \end{gathered}$$$$\begin{gathered} \mathrm{and} \mathrm{y}=-(\mathrm{x}-\mathrm{h})\mathrm{sin\theta }+(\mathrm{y}-\mathrm{k})\mathrm{cos\theta } \\ =-(1+\sqrt{2})\left(\frac{1}{\sqrt{2}}\right)+(-1-\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{1}{\sqrt{2}}\left[-1-\sqrt{2}-1-\sqrt{2}\right] \\ =\left(\frac{-2-2\sqrt{2}}{\sqrt{2}}\right) \\ =-\sqrt{2}-2 \\ Y=-2-\sqrt{2} \\ \therefore (X,Y)=(0, -2-\sqrt{2}) \end{gathered}$$