The orthocentre and the centroid of triangle | vedclass

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(C) The centroid $G$ divides the line segment joining the orthocentre $H(5,8)$ and the circumcentre $O(h,k)$ in the ratio $2:1$.Using the section form

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$4 \sqrt{10}$

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(C) The centroid $G$ divides the line segment joining the orthocentre $H(5,8)$ and the circumcentre $O(h,k)$ in the ratio $2:1$.Using the section formula,we have:$\left(\frac{2h+5}{3}, \frac{2k+8}{3}\right) = \left(3, \frac{14}{3}\right)$Equating the coordinates:$\frac{2h+5}{3} = 3$ $\Rightarrow 2h+5 = 9$ $\Rightarrow h = 2$$\frac{2k+8}{3} = \frac{14}{3}$ $\Rightarrow 2k+8 = 14$ $\Rightarrow k = 3$So,the circumcentre $O$ is $(2,3)$.The image of the orthocentre $H(5,8)$ with respect to the line $x-y=0$ (side $BC$) lies on the circumcircle. Let the image be $(x', y')$.Using the formula $\frac{x'-5}{1} = \frac{y'-8}{-1} = -2 \frac{5-8}{1^2+(-1)^2} = -2 \frac{-3}{2} = 3$.$x'-5 = 3 \Rightarrow x' = 8$$y'-8 = -3 \Rightarrow y' = 5$The point $(8,5)$ lies on the circumcircle. The radius $R$ is the distance between the circumcentre $(2,3)$ and $(8,5)$:$R = \sqrt{(8-2)^2 + (5-3)^2} = \sqrt{6^2 + 2^2} = \sqrt{36+4} = \sqrt{40} = 2\sqrt{10}$.The diameter is $2R = 2(2\sqrt{10}) = 4\sqrt{10}$.

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