In $\triangle A B C$ the mid points of the sides $A B, B C$ and $C A$ are respectively $(l, 0,0),(0, m, 0)$ and $(0,0, n)$. Then, $\frac{A B^2+B C^2+C A^2}{l^2+m^2+n^2}$ is equal to

• A

  $2$
• B

  $4$
• Correct answer
  C

  $8$
• D

  $16$

Solution

From the figure,
$$ \begin{gathered} x_1+x_2=2 l, y_1+y_2=0, z_1+z_2=0, \\ x_2+x_3=0, y_2+y_3=2 m, z_2+z_3=0 \\ \text { and } \quad x_1+x_3=0, y_1+y_3=0, z_1+z_3=2 n \end{gathered} $$ On solving, we get $$ \begin{aligned} x_1 & =l, x_2=l, x_3=-l, \\ & y_1=-m, y_2=m, y_3=m \\ \text { and } \quad z_1 & =n, z_2=-n, z_3=n \end{aligned} $$ $\therefore$ Coordinates are $A(l,-m, n), B(l, m,-n)$ and $C(-l, m, n)$ $$ \begin{aligned} \therefore & \frac{A B^2+B C^2+C A^2}{l^2+m^2+n^2} \\ = & \frac{\left(4 m^2+4 n^2\right)+\left(4 l^2+4 n^2\right)+\left(4 l^2+4 m^2\right)}{l^2+m^2+n^2} \\ = & 8 \end{aligned} $$

Hence, the correct answer is option: 'C'.