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(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$.Given $	heta = 135^o$,$x' = 4$,and $y' = -3$.The transformation fo

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$\left( \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}} \right)$

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(D) Let the original coordinates be $(x, y)$ and the new coordinates be $(x', y')$.Given $	heta = 135^o$,$x' = 4$,and $y' = -3$.The transformation formulas are:$x = x' \cos 	heta - y' \sin 	heta$$y = x' \sin 	heta + y' \cos 	heta$Substituting the values:$x = 4 \cos(135^o) - (-3) \sin(135^o) = 4 \left( -\frac{1}{\sqrt{2}} ight) + 3 \left( \frac{1}{\sqrt{2}} ight) = -\frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = -\frac{1}{\sqrt{2}}$$y = 4 \sin(135^o) + (-3) \cos(135^o) = 4 \left( \frac{1}{\sqrt{2}} ight) - 3 \left( -\frac{1}{\sqrt{2}} ight) = \frac{4}{\sqrt{2}} + \frac{3}{\sqrt{2}} = \frac{7}{\sqrt{2}}$Thus,the original coordinates are $\left( -\frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}} ight)$.

The coordinate axes are rotated through an angle of $135^o$. If the coordinates of a point $P$ in the new system are $(4, -3)$,then the coordinates of $P$ in the original system are:

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