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 $(x_1,y_1,z_1)$, E $(x_2,y_2,z_2)$ and F$(x_3,y_3,z_3)$ be the vertices of the given triangle.

And let A((-2, 3, 5), B(4, -1, 7) and C (6, 5, 3) be the mid-point 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.

$\therefore \frac{x_2+x_3}{2}=4,\frac{y_2+y_3}{2}=-1,\frac{z_2+z_3}{2}=7$

$\Rightarrow x_1+x_3=8,y_1+y_3=-2,z_1+z_3=14\left(ii\right)$

C is the mid-point of DE.

$\therefore \frac{x_2+x_3}{2}=6,\frac{y_2+y_3}{2}=5,\frac{z_2+z_3}{2}=3$

$\Rightarrow x_1+x_3=12,y_1+y_3=10,z_1+z_3=6\left(iii\right)$

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

$2(x_1+x_2+x_3)=10+14+6$

$\Rightarrow x_1+x_2+x_3=15$

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

$x_1+x_2+x_3=15$: $x_1=12,x_2=0,x_3=-4$

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

$2(y_1+y_2+y_3)=6-2+10$

$\Rightarrow y_1+y_2+y_3$ =7

Solving the last 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),:

$2(z_1+z_2+z_3)=10+14+6$

$\Rightarrow z_1+z_2+z_3=15$

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)