In a triangle PQR,the coordinates of the poin | vedclass

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(B) The circumcentre $O$ is the intersection of the perpendicular bisectors of the sides of the triangle. We are given the perpendicular bisector of $

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$(-2, -2)$

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(B) The circumcentre $O$ is the intersection of the perpendicular bisectors of the sides of the triangle. We are given the perpendicular bisector of $PR$ as $L_1: 2x - y + 2 = 0$. Next,we find the perpendicular bisector of $PQ$. The midpoint of $PQ$ is $M = (\frac{-2+4}{2}, \frac{4-2}{2}) = (1, 1)$. The slope of $PQ$ is $m_{PQ} = \frac{-2-4}{4-(-2)} = \frac{-6}{6} = -1$. The slope of the perpendicular bisector of $PQ$ is $m_{\perp} = -\frac{1}{m_{PQ}} = 1$. The equation of the perpendicular bisector of $PQ$ is $y - 1 = 1(x - 1)$,which simplifies to $y = x$ or $x - y = 0$. The circumcentre $O$ is the intersection of $2x - y + 2 = 0$ and $x - y = 0$. Substituting $y = x$ into the first equation: $2x - x + 2 = 0 \implies x = -2$. Since $y = x$,we have $y = -2$. Thus,the circumcentre is $(-2, -2)$.

In a triangle $PQR$,the coordinates of the points $P$ and $Q$ are $(-2, 4)$ and $(4, -2)$ respectively. If the equation of the perpendicular bisector of $PR$ is $2x - y + 2 = 0$,then the centre of the circumcircle of the $\Delta PQR$ is

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