Ans: B
Note that the rectangular coordinates of the image $P’$ are $(-1,-\sqrt{3})$.
Note that the rectangular coordinates of the image $P’$ are $(-1,-\sqrt{3})$.
According to the above figure, we have
$\begin{array}{rcl}
r & = & \sqrt{(1)^2 + (\sqrt{3})^2} \\
& = & 2
\end{array}$
Also,
$\begin{array}{rcl}
\tan \theta & = & \dfrac{\sqrt{3}}{1} \\
\theta & = & 60^\circ
\end{array}$
Hence, the polar coordinates of the image are $(2, 240^\circ)$.
