Question

If the mid-points of the sides AB, BC and CA of a ∆ABC are D (1, 2, –3), E (3, 0, 1) and F(–1, 1, –4) then the coordinates of the centroid of the triangle ABC are __________________.

Answer 1

Let A(x, y, z), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) be the vertices of the given triangle ABC and Let D(1, 2, –3), E(3, 0, 1) and F(–1, 1, –4) be the mid-points of sides BC, CA, and AB respectively,
figure
i.e $\frac{x_1+x_3}{2}=3$
$\frac{y_1+y_3}{2}=0\mathrm{and}\frac{z_1+z_3}{2}=1$
$$\begin{gathered} \mathrm{i}.\mathrm{e}.x_1+x_3=6 \\ {y}_1+y_3=0 \\ {z}_1+z_3=2...\left(1\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Similarly},\frac{x_1+x_2}{2}=-1,\frac{y_1+y_2}{2}=1,\frac{z_1+z_2}{2}=-4....\left(2\right) \\ \mathrm{and}\frac{x_2+x_3}{2}=1,\frac{y_2+y_3}{2}=2,\frac{z_2+z_3}{2}=-3....\left(3\right) \end{gathered}$$

i.e x₁ + x₂ = –2, y₁ + y₂ = 2, z₁ + z₂ = – 8 and x₂ + x₃ = 2, y₂ + y₃ = 4, z₂ + z₃ = – 6
∴ Centroid has co-ordinates $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3},\frac{z_1+z_2+z_3}{3}\right)$
$$\begin{gathered} \mathrm{i}.\mathrm{e}\left(\frac{2\left(x_1+x_2+x_3\right)}{6},\frac{2\left(y_1+y_2+y_3\right)}{6},\frac{2\left(z_1+z_2+z_3\right)}{6}\right) \\ \mathrm{i}.\mathrm{e}\left(\frac{x_1+x_2+x_2+x_3+x_1+x_3}{6},\frac{y_1+y_2+y_2+y_3+y_3+y_1}{6},\frac{z_1+z_2+z_2+z_3+z_3+z_1}{6}\right) \end{gathered}$$
i.e from (1), (2) and (3), we get
Coordinate of centroid
$$\begin{gathered} =\left(\frac{6-1+1}{6},\frac{0+4+2}{6},\frac{2-8-6}{6}\right) \\ =\left(1,1,-2\right) \end{gathered}$$
i.e coordinate of centroid is given by (1, 1, –2)