The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also find the centroid of the triangle.

Home » The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also find the centroid of the triangle.

CBSE | Class 11 | Introduction to Three Dimensional Geometry | Maths | NCERT Exemplar

The mid-point of the sides of a triangle are (1, 5, – 1), (0, 4, – 2) and (2, 3, 4). Find its vertices. Also find the centroid of the triangle.

Solution:

It is given that the mid-point of the sides of a triangle are $(1,5,-1),(0,4,-2)$ and $(2,3,$ , 4).
Suppose the vertices be $A\left(x_{1}, y_{1}, z_{1}\right), B\left(x_{2}, y_{2}, z_{2}\right)$ and $C\left(x_{3}, y_{3}, z_{3}\right)$ respectively.
Using midpoint formula we obtain
$\begin{array}{l} \Rightarrow\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}, \frac{z_{1}+z_{2}}{2}\right)=(1,5,-1) \\ \Rightarrow x_{1}=2-x_{2}, y_{1}=10-y_{2}, z_{1}=-2-z_{2} \\ \Rightarrow\left(\frac{x_{2}+x_{3}}{2}, \frac{y_{2}+y_{3}}{2}, \frac{z_{2}+z_{3}}{2}\right)=(0,4,-2) \\ \Rightarrow x_{3}=-x_{2}, y_{3}=8-y_{2}, z_{3}=-4-z_{2} \\ \Rightarrow\left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}, \frac{z_{1}+z_{3}}{2}\right)=(2,3,4) \\ \Rightarrow\left(\frac{2-x_{2}-x_{2}}{2}, \frac{10-y_{2}+8-y_{2}}{2}, \frac{-2-z_{2}-4-z_{2}}{2}\right)=(2,3,4) \end{array}$
On comparing the equations to their RHS we obtain
$\begin{array}{l} \therefore \mathrm{x}_{2}=-1 \\ \mathrm{y}_{2}=6 \\ \mathrm{z}_{2}=-7 \\ \therefore \mathrm{x}_{1}=2-\mathrm{x}_{2}=3 \\ \mathrm{y}_{1}=10-\mathrm{y}_{2}=4 \\ \mathrm{z}_{1}=-2-\mathrm{z}_{2}=5 \end{array}$
$\therefore \mathrm{x}_{3}=-\mathrm{x}_{2}=1$ $\mathrm{y}_{3}=8-\mathrm{y}_{2}=2$ $\mathrm{z}_{3}=-4-\mathrm{z}_{2}=3$
$\therefore \mathrm{A}(-1,6,-7), \mathrm{B}(3,4,5), \mathrm{C}(1,2,3)$ are the required vertices.
Centroid of a triangle is given by the average of the coordinates of its vertices or midpoint of sides.
Centroid is $\left(\frac{1+0+2}{3}, \frac{5+4+3}{3}, \frac{-1-2+4}{3}\right)=(1,4,1 / 3)$