Question

In $△ABC$ the mid points of the sides $AB,BC$ and $CA$ are respectively $(l,0,0),(0,m,0)$ and $(0,0,n)$. Then, $\frac{{AB}^2+{BC}^2+{CA}^2}{l^2+m^2+n^2}$ is equal to

• A. $2$
• B. $4$
• C. $8$
• D. $16$

Answer 1

The correct option is C $8$

Let $A=(x_1,y_1,z_1),B=(x_2,y_2,z_2),C=(x_3,y_3,z_3)$

From the figure,
$x_1+x_2=2l,y_1+y_2=0,z_1+z_2=0$, [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
$x_1=l,x_2=l,x_3=-l$,

$y_1=-m,y_2=m,y_3=m$

and $z_1=n,z_2=-n,z_3=n$

$\therefore$ Coordinates are $A(l,-m,n),B(l,m,-n)$ and $C(-l,m,n)$

$\therefore \frac{{AB}^2+{BC}^2+{CA}^2}{l^2+m^2+n^2}$

$=\frac{4m^2+4n^2+4l^2+4n^2+(4l^2+4m^2)}{l^2+m^2+n^2}$

$=8$