question_answer

In \[\Delta ABC\] the mid-point 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

Correct Answer: C

Solution :

[c] From the figure

\[{{x}_{1}}+{{x}_{2}}=21,{{y}_{1}}+{{y}_{2}}=0,{{z}_{1}}+{{z}_{2}}=0\]

\[{{x}_{2}}+{{x}_{3}}=0,{{y}_{2}}+{{y}_{3}}=2m,{{z}_{1}}+{{z}_{3}}=0\]

and \[{{x}_{1}}+{{x}_{3}}=0,{{y}_{1}}+{{y}_{3}}=0,{{z}_{1}}+{{z}_{3}}=2n\]

On solving, we get \[{{x}_{1}}=l,{{x}_{2}}=l,{{x}_{3}}=-1\]

\[{{y}_{1}}=-m,{{y}_{2}}=m,{{y}_{3}}=m\] and \[{{z}_{1}}=n,{{z}_{2}}=-n,{{z}_{3}}=n\]

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

\[\therefore \frac{A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}}{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}\]

\[=\frac{(4{{m}^{2}}+4{{n}^{2}})+(4{{l}^{2}}+4{{n}^{2}})+(4{{l}^{2}}+4{{m}^{2}})}{{{l}^{2}}+{{m}^{2}}+{{n}^{2}}}=8\]