Question

In $\Delta ABC$ the mid points of the sides $AB,BC$ and $CA$ 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,

$x_1+x_2=2l,y_1+y_2=0,z_1+z_2=0$,
$x_2+x_3=0,y_2+y_3=2m,z_2+z_3=0$
and $x_1+x_3=0,y_1+y_3=0,z_1+z_3=2n$
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 $C(-l,m,n)$
$\therefore \frac{A{B}^2+B{C}^2+C{A}^2}{l^2+m^2+n^2}$
$=\frac{\left(4m^2+4n^2\right)+\left(4l^2+4n^2\right)+\left(4l^2+4m^2\right)}{l^2+m^2+n^2}$
$=8$