The point P(3,6) is first reflected on the li | vedclass

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(D) The reflection of point $P(3,6)$ on the line $y=x$ gives the point $Q(6,3)$.The reflection of point $Q(6,3)$ on the line $y=-x$ gives the point $Q

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$(0,0)$

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(D) The reflection of point $P(3,6)$ on the line $y=x$ gives the point $Q(6,3)$.The reflection of point $Q(6,3)$ on the line $y=-x$ gives the point $Q^{\prime}(-3,-6)$.Now, the slope of $PQ = \frac{3-6}{6-3} = \frac{-3}{3} = -1$.The slope of $QQ^{\prime} = \frac{-6-3}{-3-6} = \frac{-9}{-9} = 1$.Since the product of the slopes is $(-1) 	imes (1) = -1$, the lines $PQ$ and $QQ^{\prime}$ are perpendicular.Therefore, $\Delta PQQ^{\prime}$ is a right-angled triangle with the right angle at $Q$.The circumcentre of a right-angled triangle is the midpoint of its hypotenuse.The hypotenuse is $PQ^{\prime}$, with endpoints $P(3,6)$ and $Q^{\prime}(-3,-6)$.Midpoint of $PQ^{\prime} = \left(\frac{3+(-3)}{2}, \frac{6+(-6)}{2}ight) = (0,0)$.

The point $P(3,6)$ is first reflected on the line $y=x$ and then the image point $Q$ is again reflected on the line $y=-x$ to get the image point $Q^{\prime}$. Then, the circumcentre of the $\Delta P Q Q^{\prime}$ is

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