If a vertex of a triangle is $(1, 1)$ and the mid points of two sides through this vertex are $(- 1, 2), (3, 2)$ , then the centroid of the triangle is

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(\frac{1}{3}, \frac{7}{3})

(\frac{- 1}{3}, \frac{7}{3})

(- 1, \frac{7}{3})

(1, \frac{7}{3})

answer is C.

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Detailed Solution

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Let $A = (1, 1)$

Mid point of $A B = (- 1, 2)$

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (- 1, 2)$

$(\frac{1 + x}{2}, \frac{1 + y}{2}) = (- 1, 2)$

$\frac{1 + x}{2} = - 1$ , $\frac{1 + y}{2} = 2$

$1 + x = - 2$ , $1 + y = 4$

$x = - 3$ , $y = 3$

$∴$ point $B = (- 3, 3)$

Mid point of $A C = (3, 2)$

$(\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2}) = (3, 2)$

$(\frac{a + 1}{2}, \frac{b + 1}{2}) = (3, 2)$

$\frac{a + 1}{2} = 3$ , $\frac{b + 1}{2} = 2$

$a + 1 = 6$ , $b + 1 = 4$

$a = 5$ , $b = 3$

$∴$ point $C = (5, 3)$

Centroid of $Δ A B C = [\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}]$

$= [\frac{1 - 3 + 5}{3}, \frac{1 + 3 + 3}{3}]$

$= (\frac{3}{3}, \frac{7}{3})$

$G = (1, \frac{7}{3})$

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