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 $(0,-2-\sqrt{2})$.