Question

The mid-points of the sides of a triangle ABC are given by (–2, 3, 5), (4, –1, 7) and (6, 5, 3). Find the coordinates of A, B and C.

Answer 1

Let\[D \left( x_1 , y_1 , z_1 \right), E \left( x_2 , y_2 , z_2 \right) \text{ and } F \left( x_3 , y_3 , z_3 \right)\]  be the vertices of the given triangle.
And, let\[A \left( - 2, 3, 5 \right), B \left( 4, - 1, 7 \right) \text{ and } C \left( 6, 5, 3 \right)\] 

be the mid-points of the sides EF, FA and DE, respectively.
Now, A is the mid-point of EF.

\[\therefore \frac{x_2 + x_3}{2} = - 2, \frac{y_2 + y_3}{2} = 3, \frac{z_2 + z_3}{2} = 5\]
\[ \Rightarrow x_2 + x_3 = - 4, y_2 + y_3 = 6, z_2 + z_3 = 10 \left( i \right)\]

B is the mid-point of FD. 

\[\therefore \frac{x_1 + x_3}{2} = 4, \frac{y_1 + y_3}{2} = - 1, \frac{z_1 + z_3}{2} = 7\]
\[ \Rightarrow x_1 + x_3 = 8, y_1 + y_3 = - 2, z_1 + z_3 = 14 \left( ii \right)\]

C is the mid-point of DE.

\[\therefore \frac{x_1 + x_2}{2} = 6, \frac{y_1 + y_2}{2} = 5, \frac{z_1 + z_2}{2} = 3\]
\[ \Rightarrow x_1 + x_2 = 12, y_1 + y_2 = 10, z_1 + z_2 = 6 \left( iii \right)\]

Adding the first three equations in (i), (ii) and (iii): 

\[2\left( x_1 + x_2 + x_3 \right) = - 4 + 8 + 12\]
\[ \Rightarrow x_1 + x_2 + x_3 = 8\]

Solving the first three equations in (i), (ii) and (iii) with \[x_1 + x_2 + x_3 = 8\]

\[x_1 = 12, x_2 = 0, x_3 = - 4\]

Adding the next three equations in (i), (ii) and (iii):

\[2\left( y_1 + y_2 + y_3 \right) = 6 - 2 + 10\]
\[ \Rightarrow y_1 + y_2 + y_3 = 7\]

Solving the next three equations in (i), (ii) and (iii) with \[y_1 + y_2 + y_3 = 7\] 

\[y_1 = 1, y_2 = 9, y_3 = - 3\]

Adding the last three equations in (i), (ii) and (iii):

\[2\left( z_1 + z_2 + z_3 \right) = 10 + 14 + 6\]
\[ \Rightarrow z_1 + z_2 + z_3 = 15\] 

Solving the last three equations in (i), (ii) and (iii) with

\[z_1 + z_2 + z_3 = 15\]

\[z_1 = 5, z_2 = 1, z_3 = 9\]

Thus, the vertices of the triangle are (12, 1, 5), (0, 9, 1), (−4, −3, 9).