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Q.
The mid-points of the sides of a triangle are $(5,7,11),(0,8,5)$ and $(2,3,-1)$, then the vertices are
Introduction to Three Dimensional Geometry
Solution:
Let the vertices of a triangle are $A\left(x_1, y_1, z_1\right)$,
$B\left(x_2, y_2, z_2\right) \text { and } C\left(x_3, y_3, z_3\right)$
Since, $D, E$ and $F$ are the mid-points of $A C, B C$ and $A B$.
$\therefore \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)=(0,8,5) $
$\Rightarrow x_1+x_2=0, y_1+y_2=16, z_1+z_2=10 ....$(i)
$\Rightarrow \left(\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2}, \frac{z_2+z_3}{2}\right)=(2,3,-1) $
$\Rightarrow x_2 + x_3 = 4, y_2 + y_3 = 6, z_2 + z_3 = -2.....$(ii(
and $ x_2+x_3=4, y_2+y_3=6, z_2+z_3=-2$
$\Rightarrow \left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}, \frac{z_1+z_3}{2}\right)=(5,7,11) $
$\Rightarrow x_1+x_3=10, y_1+y_3=14, z_1+z_3=22 ....$(iii)
On adding Eqs. (i), (ii) and (iii), we get
$2\left(x_1+x_2+x_3\right)=14,2\left(y_1+y_2+y_3\right) =36, $
$2\left(z_1+z_2+z_3\right) =30$
$\Rightarrow x_1+x_2+x_3=7, y_1+y_2+y_3 =18$
$z_1+z_2+z_3 =15 ....$(iv)
On solving Eqs. (i), (ii), (iii) and (iv), we get
$x_3=7, x_1=3, x_2=-3$
$ y_3=2, y_1=12, y_2=4$
and $z_3=5, z_1=17, z_2=-7$
Hence, vertices of a triangle are $(7,2,5),(3,12,17)$ and $(-3,4,-7)$.
