Question

If $BC,CA$ and $AB$ are the sides of a triangle $ABC$ whose midpoints are $(p,0,0),(0,q,0),(0,0,r)$ then find $\frac{(AB{)}^2+(BC{)}^2+(CA{)}^2}{p^2+q^2+r^2}$

• A. $8$
• B. $6$
• C. $5$
• D. $2$

Answer 1

The correct option is D $2$

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

$M=\frac{x_2+x_3}{2},\frac{y_2+y_3}{2},\frac{z_2+z_3}{2}$

$\Rightarrow x_2+x_3=p$, $y_2+y_3=0$, $z_2+z_3=0$

$N=\frac{x_1+x_3}{2},\frac{y_1+y_3}{2},\frac{z_1+z_3}{2}$

$\Rightarrow x_1+x_3=0$, $y_1+y_3=9$, $z_1+z_3=0$

$O=\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2}\Rightarrow x_1+x_2=0$, $y_1+y_2=0$, $z_1+z_2=r$

Solving,

$x_2=p/2$, $x_1=-p/2$, $x_3=p/2$

$y_3=q/2$, $y_1=q/2$, $y_2=-q/2$

$z_3=-r/2$, $z_1=r/2$, $z_2=r/2$

$\therefore$ $A=(-p/2,q/2,r/2)$ ; $B=(p/2,-q/2,r/2)$ ; $C=(p/2,q/2,-r/2)$

${AB}^2=p^2+q^2$ ${AC}^2=p^2+r^2$

${BC}^2=q^2+r^2$

$\therefore$ $\frac{{AB}^2+{BC}^2+{CA}^2}{p^2+q^2+r^2}=2$ [D]