Find the corrdinates of the point of intersection of the medians of triangle ABC; given A $= (- 2, 3)$ , B $= (6, 7)$ and C $= (4, 1)$ .

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Solution

For the vertices A $(x_{1}, y_{1})$ , B $(x_{2}, y_{2})$ and C $(x_{3}, y_{3})$ of triangle ABC, its centroid $= (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3})$
Now, taking $(- 2, 3) = (x_{1}, y_{1})$ , $(6, 7) = (x_{2}, y_{2})$ and $(4, 1) = (x_{3}, y_{3})$ , we get
Thus, centroid $= (\frac{- 2 + 6 + 4}{3}, \frac{3 + 7 + 1}{3})$
$= (\frac{8}{3}, \frac{11}{3})$

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Let ABC be a triangle with altitudes h1,h2,h3 and inradius r. Find the minimum value of the expression

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