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If the origin of a coordinate system is shifted to $(-\sqrt{2}, \sqrt{2})$ and the coordinate system is rotated anti-clockwise through an angle $45^{\circ}$, then the point $P(1,-1)$ in the original system has new coordinates
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Verified Answer
The correct answer is:
$(0,-2-\sqrt{2})$
Let the new coordinates of the point be $(X, Y)$.
Since, the origin is shifted to $(-\sqrt{2}, \sqrt{2})$ and rotates anti-clockwise through an angle $45^{\circ}$.
$$
\begin{aligned}
& \text { So, }(h, k)=(-\sqrt{2}, \sqrt{2}), \theta=\frac{\pi}{4} \text { and }(x, y)=(1,-1) \\
& \therefore \quad X=(x-h) \cos \theta+(y-k) \sin \theta \\
& =(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=0 \\
& \text { and } \quad Y=-(x-h) \sin \theta+(y-k) \cos \theta \\
& =-(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=-2-\sqrt{2} \\
&
\end{aligned}
$$
Hence, the new coordinates are $(0,-2-\sqrt{2})$.
Since, the origin is shifted to $(-\sqrt{2}, \sqrt{2})$ and rotates anti-clockwise through an angle $45^{\circ}$.
$$
\begin{aligned}
& \text { So, }(h, k)=(-\sqrt{2}, \sqrt{2}), \theta=\frac{\pi}{4} \text { and }(x, y)=(1,-1) \\
& \therefore \quad X=(x-h) \cos \theta+(y-k) \sin \theta \\
& =(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=0 \\
& \text { and } \quad Y=-(x-h) \sin \theta+(y-k) \cos \theta \\
& =-(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=-2-\sqrt{2} \\
&
\end{aligned}
$$
Hence, the new coordinates are $(0,-2-\sqrt{2})$.
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