If in triangle ABC,A equiv (1, 10),circumcent | vedclass

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(A) We know that the orthocentre $(H)$,centroid $(G)$,and circumcentre $(O)$ of any triangle are collinear,and the centroid $G$ divides the line segme

Vedclass · Accepted Answer

$(1, -11/3)$

Vedclass · Answer

(A) We know that the orthocentre $(H)$,centroid $(G)$,and circumcentre $(O)$ of any triangle are collinear,and the centroid $G$ divides the line segment $HO$ in the ratio $2:1$.Let $H = (\frac{11}{3}, \frac{4}{3})$ and $O = (-\frac{1}{3}, \frac{2}{3})$.The coordinates of the centroid $G$ are given by the section formula:$G = \left( \frac{2(-\frac{1}{3}) + 1(\frac{11}{3})}{2+1}, \frac{2(\frac{2}{3}) + 1(\frac{4}{3})}{2+1} \right) = \left( \frac{-\frac{2}{3} + \frac{11}{3}}{3}, \frac{\frac{4}{3} + \frac{4}{3}}{3} \right) = \left( \frac{3}{3}, \frac{8/3}{3} \right) = \left( 1, \frac{8}{9} \right)$.Let $D(h, k)$ be the mid-point of the side opposite to $A$. The centroid $G$ divides the median $AD$ in the ratio $2:1$.Using the section formula for $G$ on $AD$ where $A = (1, 10)$ and $D = (h, k)$:$1 = \frac{2h + 1(1)}{2+1} \implies 3 = 2h + 1 \implies 2h = 2 \implies h = 1$.$\frac{8}{9} = \frac{2k + 1(10)}{2+1} \implies \frac{8}{9} = \frac{2k + 10}{3} \implies \frac{8}{3} = 2k + 10 \implies 2k = \frac{8}{3} - 10 = \frac{8-30}{3} = -\frac{22}{3} \implies k = -\frac{11}{3}$.Thus,the coordinates of the mid-point $D$ are $(1, -11/3)$.

If in triangle $ABC$,$A \equiv (1, 10)$,circumcentre $\equiv \left( -\frac{1}{3}, \frac{2}{3} \right)$ and orthocentre $\equiv \left( \frac{11}{3}, \frac{4}{3} \right)$,then the coordinates of the mid-point of the side opposite to $A$ are:

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