If the axes are rotated through an angle 45^{ | vedclass

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(A) Let the old coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The transformation equations for rotation of axes by an angle $	heta$

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$(1-2 \sqrt{2}, 1+2 \sqrt{2})$

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(A) Let the old coordinates be $(x, y)$ and the new coordinates be $(x', y')$. The transformation equations for rotation of axes by an angle $	heta$ are:$x = x' \cos 	heta - y' \sin 	heta$$y = x' \sin 	heta + y' \cos 	heta$Given $	heta = 45^{\circ}$,$x' = \sqrt{2}$,and $y' = 4$.Substituting these values:$x = \sqrt{2} \cos 45^{\circ} - 4 \sin 45^{\circ}$$x = \sqrt{2} \left( \frac{1}{\sqrt{2}} ight) - 4 \left( \frac{1}{\sqrt{2}} ight) = 1 - \frac{4}{\sqrt{2}} = 1 - 2\sqrt{2}$$y = \sqrt{2} \sin 45^{\circ} + 4 \cos 45^{\circ}$$y = \sqrt{2} \left( \frac{1}{\sqrt{2}} ight) + 4 \left( \frac{1}{\sqrt{2}} ight) = 1 + \frac{4}{\sqrt{2}} = 1 + 2\sqrt{2}$Thus,the coordinates in the old system are $(1 - 2\sqrt{2}, 1 + 2\sqrt{2})$.

If the axes are rotated through an angle $45^{\circ}$ in the positive direction without changing the origin,then the coordinates of the point $(\sqrt{2}, 4)$ in the old system are

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