Question

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

• A. $(\sqrt{2},-2 \sqrt{2})$
• B. $(0,-2 \sqrt{2})$
• C. $(0,-2-\sqrt{2})$
• D. $(0,-2+\sqrt{2})$

Answer 1

Answer: (C)

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 $45^{\circ}$.
So,$(h, k)=(-\sqrt{2}, \sqrt{2}), \theta=\frac{\pi}{4} \text { and }(x, y)=(1,-1)$
$\therefore X=(x-h) \cos \theta+(y-k) \sin \theta $
$=(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=0 $
and $Y=-(x-h) \sin \theta+(y-k) \cos \theta $
$=-(1+\sqrt{2}) \frac{1}{\sqrt{2}}+(-1-\sqrt{2}) \frac{1}{\sqrt{2}}=-2-\sqrt{2}$
Hence, the new coordinates are $(0,-2-\sqrt{2})$.