Question

The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4). Find its vertices.

Answer 1

Let $A(x_1,y_1,z_1),B(x_2,y_2,z_2)$ and $C(x_3,y_3,z_3)$ be the vertices of the given triangle, and let D(1, 5, -1), E(0, 4, -2) and F(2, 3, 4) be the midpoints of the sides BC, CA and AB respectively. Then,

$\frac{x_2+x_3}{2}=1;\frac{y_2+y_3}{2}=5;\frac{z_2+z_3}{2}=1;$

$\frac{x_3+x_1}{2}=0;\frac{y_3+y_1}{2}=4;\frac{z_3+z_1}{2}=-2;$

$\frac{x_1+x_2}{2}=2;\frac{y_1+y_2}{2}=3$ and $\frac{z_1+z_2}{2}=4$.

Thus, $x_2+x_3=2;x_3+x_1=0;x_1+x_2=4;$

$y_2+y_3=10;y_3+y_1=8;y_1+y_2=6;$

$z_2+z_3=-2;z_3+z_1=-4;z_1+z_2=8$.

Adding first three equations , we get

$2(x_1+x_2+x_3)=6$ or $x_1+x_2+x_3=3$.

Thus, $x_1=1,x_2=3$ and $x_3=-1$.

Adding next three equations, we get

$2(y_1+y_2+y_3)=24$ or $y_1+y_2+y_3=12$.

$\therefore y_1=2;y_2=4$ and $y_3=6$.

Adding last three equations, we get

$2(z_1+z_2+z_3)=2$ or $z_1+z_2+z_3=1$

$\therefore z_1=3,z_2=5$ and $z_3=-7$.

Hence, the vertices of the given triangle are

A(1,2,3); B(3,4,5) and C(-1,6,-7).