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 $D\left(x_1,y_1,z_1\right),E\left(x_2,y_2,z_2\right)\mathrm{and}F\left(x_3,y_3,z_3\right)$ be the vertices of the given triangle.
And, let $A\left(-2,3,5\right),B\left(4,-1,7\right)\mathrm{and}C\left(6,5,3\right)$ be the mid-points of the sides EF, FA and DE, respectively.
Now, A is the mid-point of EF.
$$\begin{gathered} \therefore \frac{x_2+x_3}{2}=-2,\frac{y_2+y_3}{2}=3,\frac{z_2+z_3}{2}=5 \\ \Rightarrow x_2+x_3=-4,y_2+y_3=6,z_2+z_3=10\left(i\right) \end{gathered}$$
B is the mid-point of FD.
$$\begin{gathered} \therefore \frac{x_1+x_3}{2}=4,\frac{y_1+y_3}{2}=-1,\frac{z_1+z_3}{2}=7 \\ \Rightarrow x_1+x_3=8,y_1+y_3=-2,z_1+z_3=14\left(ii\right) \end{gathered}$$

C is the mid-point of DE.
$$\begin{gathered} \therefore \frac{x_1+x_2}{2}=6,\frac{y_1+y_2}{2}=5,\frac{z_1+z_2}{2}=3 \\ \Rightarrow x_1+x_2=12,y_1+y_2=10,z_1+z_2=6\left(iii\right) \end{gathered}$$
Adding the first three equations in (i), (ii) and (iii):
$$\begin{gathered} 2\left(x_1+x_2+x_3\right)=-4+8+12 \\ \Rightarrow x_1+x_2+x_3=8 \end{gathered}$$
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):
$$\begin{gathered} 2\left(y_1+y_2+y_3\right)=6-2+10 \\ \Rightarrow y_1+y_2+y_3=7 \end{gathered}$$
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):
$$\begin{gathered} 2\left(z_1+z_2+z_3\right)=10+14+6 \\ \Rightarrow z_1+z_2+z_3=15 \end{gathered}$$
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).