Question

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.

Answer 1

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)

\[\frac{x_1 + x_2}{2} = 1, \frac{y_1 + y_2}{2} = 2 \text{ and } \frac{z_1 + z_2}{2} = - 3\]
\[ \Rightarrow x_1 + x_2 = 2, y_1 + y_2 = 4 \text{ and } z_1 + z_2 = - 6 . . . . . \left( 1 \right)\]

Mid point of BC is E(3, 0,1)

\[\frac{x_2 + x_3}{2} = 3, \frac{y_2 + y_3}{2} = 0 \text{ and } \frac{z_2 + z_3}{2} = 1\]
\[ \Rightarrow x_2 + x_3 = 6, y_2 + y_3 = 0 \text{ and } z_2 + z_3 = 2 . . . . . \left( 2 \right)\]

Mid point of AC is F(−1, 1, −4)

\[\frac{x_1 + x_3}{2} = - 1, \frac{y_1 + y_3}{2} = 1 \text{ and } \frac{z_1 + z_3}{2} = - 4\]
\[ \Rightarrow x_1 + x_3 = - 2, y_1 + y_3 = 2 \text{ and } z_1 + z_3 = - 8 . . . . . \left( 3 \right)\]

Adding (1), (2) and (3),we get 

\[2\left( x_1 + x_2 + x_3 \right) = 6, 2\left( y_1 + y_2 + y_3 \right) = 6 \text{ and } 2\left( z_1 + z_2 + z_3 \right) = - 12\]
\[ \Rightarrow x_1 + x_2 + x_3 = 3, y_1 + y_2 + y_3 = 3 \text{ and } z_1 + z_2 + z_3 = - 6\]
\[ \Rightarrow \frac{x_1 + x_2 + x_3}{3} = 1, \frac{y_1 + y_2 + y_3}{3} = 1 \text{ and } \frac{z_1 + z_2 + z_3}{3} = - 2\]
Hence, the centroid of the traingle ABC is (1, 1, −2).