Question

In $△ABC$ the mid points of the sides $AB,BC$ and $CA$ are respectively $(1,0,0),(0$, $m,0)$ 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)

Let $A=\left(x_1,y_1,z_1\right),$
$B=\left(x_2,y_2,z_2\right),$
$C=\left(x_3,y_3,z_3\right)$
From the figure,
$x_1+x_2=21,y_1+y_2=0,z_1+z_2=0,$ [midpoint formula]
$x_2+x_3=0,y_2+y_3=2m,z_2+z_3=0$
and $x_1+x_3=0,y_1+y_3=0,z_1+z_3=2n$
On solving, we get
$x_1=1,x_2=1,x_3=-1$,
$y_1=-m,y_2=m,y_3=m$
and $z_1=n,z_2=-n,z_3=n$
$\therefore$ Coordinates are $A(1,-m,n),B(1,m,-n)$ and $C(-1,m,n)$
$\therefore \frac{A{B}^2+B{C}^2+C{A}^2}{1^2+m^2+n^2}$
$=\frac{4m^2+4n^2+41^2+4n^2+\left(41^2+4m^2\right)}{1^2+m^2+n^2}$
$=8$