If the coordinates of a point P changes to (2, –6) when the coordinate axes are rotated through an angle of ${135}^0$, then the coordinates of P in the original system are

$$\begin{gathered} \mathrm{Given} \mathrm{\theta }={135}^0 \\ \mathrm{Let} (\mathrm{x},\mathrm{y}) \mathrm{be} \mathrm{original} \mathrm{coordinates} \\ \mathrm{Given} (\mathrm{x},\mathrm{y})=(2,-6) \\ \mathrm{Now} \mathrm{x}=X\mathrm{cos\theta }-Y \mathrm{sin\theta } \\ =2 \cos {135}^0+6 \sin {135}^0 \\ \mathrm{x}=2\left(\frac{-1}{\sqrt{2}}\right)+6\left(\frac{1}{\sqrt{2}}\right) \\ =\frac{4}{\sqrt{2}} \\ \mathrm{x}=2\sqrt{2} \\ \mathrm{and} \mathrm{y}=\mathrm{Xsin\theta }+Y \mathrm{cos\theta } \\ =2 \sin {135}^0-6 \cos {135}^0 \\ =2\left(\frac{1}{\sqrt{2}}\right)-6(-\frac{1}{\sqrt{2}}) \\ =\frac{2}{\sqrt{2}}+\frac{6}{\sqrt{2}} \\ =\frac{8}{\sqrt{2}} \\ =4\sqrt{2} \\ \therefore \mathrm{Original} \mathrm{coordinates} (\mathrm{x},\mathrm{y})=(2\sqrt{2},4\sqrt{2}) \end{gathered}$$