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  4. In a triangle A B C, if the mid-points of sides A B, B C, C A are (3,0,0),(0,4,0),(0,0,5) respectively, then A B2+B C2+C A2=

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Q. In a triangle $A B C$, if the mid-points of sides $A B, B C, C A$ are $(3,0,0),(0,4,0),(0,0,5)$ respectively, then $A B^{2}+B C^{2}+C A^{2}=$

AP EAMCETAP EAMCET 2019

Solution:

Given, mid-point of sides $AB, BC, CA$ are
$(3, 0, 0), (0, 4, 0), (0, 0, 5)$.
image
$\therefore A =(3,-4,5) $
$ B =(3,4,-5) $ and $ C=(-3,4,5) $
$A B^{2} =\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2} $
$=(3-3)^{2}+(4+4)^{2}+(-5-5)^{2} $
$=8^{2}+10^{2}=164 $
$ B C^{2} =(-3-3)^{2}+(4-4)^{2}+(5+5)^{2}$
$=36+100=136$
And $ C A^{2}=(-3-3)^{2}+(4+4)^{2}+(5-5)^{2}$
$= (-6)^{2}+(8)^{2}=36+64=100 $
Now, $A B^{2}+B C^{2}+C A^{2}=164+136+100=400$