In any triangle, the centroid G divides the line segment joining the orthocentre H((11)/(3), (4)/(3)) and the circumcentre O(-(1)/(3), (2)/(3)) internally in the ratio 2: 1. Applying the section formula gives x_G = (-(2)/(3) + (11)/(3))/(3) = 1 and y_G = ((4)/(3) + (4)/(3))/(3) = (8)/(9).

In any triangle, the centroid G divides the line segment joining the orthocentre H((11)/(3), (4)/(3)) and the circumcentre O(-(1)/(3), (2)/(3)) internally in the ratio 2: 1. Applying the section formula gives x_G = (-(2)/(3) + (11)/(3))/(3) = 1 and y_G = ((4)/(3) + (4)/(3))/(3) = (8)/(9).

Let D be the mid-point of the side BC opposite to vertex A(1, 10). Since the centroid G divides the median AD in the ratio 2: 1, we have G = (A + 2D)/(3). Solving for D gives x_D = (3(1) - 1)/(2) = 1 and y_D = (3((8)/(9)) - 10)/(2) = ((8)/(3) - 10)/(2) = -(11)/(3). Hence, option D is correct.

Let D be the mid-point of the side BC opposite to vertex A(1, 10). Since the centroid G divides the median AD in the ratio 2: 1, we have G = (A + 2D)/(3). Solving for D gives x_D = (3(1) - 1)/(2) = 1 and y_D = (3((8)/(9)) - 10)/(2) = ((8)/(3) - 10)/(2) = -(11)/(3). Hence, option D is correct.

Problem 1: The co-ordinates of mid points of the sides of a triangle are $(4, 2)$ , $(3, 3)$ and $(2, 2)$ . Then the coordinates of the centroid of the triangle are Options: A $\left(3, \frac{7}{3}\right)$ B $(3, 3)$ C $(4, 3)$ D $(1, 2)$ --- Problem 2: If in a triangle $A = (1, 10)$ , circumcentre $= \left(-\frac{1}{3}, \frac{2}{3}\right)$ and orthocentre $\left(\frac{11}{3}, \frac{4}{3}\right)$ , then the co-ordinate of mid-point of side opposite to $A$ is Options: A $(1, 6)$ B $(1, 5)$ C $(1, -3)$ D $\left(1, -\frac{11}{3}\right)$

Answer: Problem 1: (A) $\left(3, \frac{7}{3}\right)$ ; Problem 2: (D) $\left(1, -\frac{11}{3}\right)$

Step-by-step solution

Step 1: Find the centroid in Problem 1

A fundamental property in coordinate geometry states that the centroid of a triangle coincides with the centroid of the medial triangle formed by its mid-points. Substituting the mid-point coordinates $(4, 2)$ , $(3, 3)$ , and $(2, 2)$ into the centroid formula gives $G = \left(3, \frac{7}{3}\right)$ . Hence, option A is correct for Problem 1.

Step 2: Find the centroid using Euler line ratio in Problem 2

In any triangle, the centroid $G$ divides the line segment joining the orthocentre $H\left(\frac{11}{3}, \frac{4}{3}\right)$ and the circumcentre $O\left(-\frac{1}{3}, \frac{2}{3}\right)$ internally in the ratio $2 : 1$ . Applying the section formula gives $x_G = \frac{-\frac{2}{3} + \frac{11}{3}}{3} = 1$ and $y_G = \frac{\frac{4}{3} + \frac{4}{3}}{3} = \frac{8}{9}$ .

Step 3: Determine the mid-point of the opposite side in Problem 2

Let $D$ be the mid-point of the side $BC$ opposite to vertex $A(1, 10)$ . Since the centroid $G$ divides the median $AD$ in the ratio $2 : 1$ , we have $G = \frac{A + 2D}{3}$ . Solving for $D$ gives $x_D = \frac{3(1) - 1}{2} = 1$ and $y_D = \frac{3\left(\frac{8}{9}\right) - 10}{2} = \frac{\frac{8}{3} - 10}{2} = -\frac{11}{3}$ . Hence, option D is correct.

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