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If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid

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#### Solution

Let P (1, 1), Q(2, –3), R(3, 4) be the mid-points of sides AB, BC and CA respectively of triangle ABC. Let `A(x_1 , y_1 ), B(x_2 , y_2 ) ` be the vertices of triangle ABC.

Then, P is the mid-point of BC

`\Rightarrow (x_{1}+x_{2})/{2}=1,(y_{1}+y_{2})/{2}=1`

`⇒ x_1 + x_2 = 2 and y_1 + y_2 = 2 …(1)`

Q is the mid-point of BC

`\Rightarrow (x_{2}+x_{3})/{2}=2,(y_{2}+y_{3})/{2}=-3`

`⇒ x_2 + x_3 = 4 and y_2 + y_3 = – 6 …(2)`

R is the mid-point of AC

`\Rightarrow (x_{1}+x_{3})/{2}=3,(y_{1}+y_{3})/{2}=4`

`⇒ x_1 + x_3 = 6 and y1_1 + y_3 = 8 …(3)`

From (1), (2) and (3), we get

`x_1 + x_2 + x_2 + x_3 + x_1 + x_3 = 2 + 4 + 6 and, y_1 + y_2 + y_2 + y_3 + y_1 + y_3 = 2 – 6 + 8`

`x_1 + x_2 + x_3 = 6 and y_1 + y_2 + y_3 = 2 …(4)`

The coordinates of the centroid of ∆ABC are

`( (x_{1}+x_{2}+x_{3})/{3},(y_{1}+y_{2}+y_{3})/{3})=( \frac{6}{3},\frac{2}{3})`

`=( 2,\ \frac{2}{3})`

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