The coordinates of the mid-points of sides AB, BC and CA of △ABC are D(1, 2, −3), E(3, 0,1) and F(−1, 1, −4) respectively. Write the coordinates of its centroid.

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Solution

Let the coordintes of the triangles be A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃).
Now,
Mid point of AB is D(1, 2, −3)
$\frac{x_{1} + x_{2}}{2} = 1, \frac{y_{1} + y_{2}}{2} = 2 and \frac{z_{1} + z_{2}}{2} = - 3 \Rightarrow x_{1} + x_{2} = 2, y_{1} + y_{2} = 4 and z_{1} + z_{2} = - 6 . . . . . (1)$

Mid point of BC is E(3, 0,1)
$\frac{x_{2} + x_{3}}{2} = 3, \frac{y_{2} + y_{3}}{2} = 0 and \frac{z_{2} + z_{3}}{2} = 1 \Rightarrow x_{2} + x_{3} = 6, y_{2} + y_{3} = 0 and z_{2} + z_{3} = 2 . . . . . (2)$

Mid point of AC is F(−1, 1, −4)

$\frac{x_{1} + x_{3}}{2} = - 1, \frac{y_{1} + y_{3}}{2} = 1 and \frac{z_{1} + z_{3}}{2} = - 4 \Rightarrow x_{1} + x_{3} = - 2, y_{1} + y_{3} = 2 and z_{1} + z_{3} = - 8 . . . . . (3)$

Adding (1), (2) and (3),we get
$2 (x_{1} + x_{2} + x_{3}) = 6, 2 (y_{1} + y_{2} + y_{3}) = 6 and 2 (z_{1} + z_{2} + z_{3}) = - 12 \Rightarrow x_{1} + x_{2} + x_{3} = 3, y_{1} + y_{2} + y_{3} = 3 and z_{1} + z_{2} + z_{3} = - 6 \Rightarrow \frac{x_{1} + x_{2} + x_{3}}{3} = 1, \frac{y_{1} + y_{2} + y_{3}}{3} = 1 and \frac{z_{1} + z_{2} + z_{3}}{3} = - 2$
Hence, the centroid of the traingle ABC is (1, 1, −2).

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