Question

In $\Delta A B C$ 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
• C. 8
• D. 16

Answer 1

Answer: (C)

From the figure,

$x_{1}+x_{2}=2 l, y_{1}+y_{2}=0, z_{1}+z_{2}=0$,
$x_{2}+x_{3}=0, y_{2}+y_{3}=2 m, z_{2}+z_{3}=0$
and $x_{1}+x_{3}=0, y_{1}+y_{3}=0, z_{1}+z_{3}=2 n$
On solving, we get
$x_{1}=l, x_{2}=l, x_{3}=-l$,
and $y_{1}=-m, y_{2}=m, y_{3}=m$
$z_{1}=n, z_{2}=-n, z_{3}=n$
$\therefore $ Coordinates are $A(l,-m, n), B(l, m,-n)$ and $C(-l, m, n)$
$ \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 $