Q60 / 160 · TS EAMCET 2015 · Mathematics

$\frac{1}{r^2}+\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2}$ equals

  • A
    $\frac{a^2+b^2+c^2}{\Delta}$
  • Correct answer
    B
    $\frac{a^2+b^2+c^2}{\Delta^2}$
  • C
    $\frac{\Delta^2}{a^2+b^2+c^2}$
  • D
    $\frac{\Delta}{a^2+b^2+c^2}$

Solution

We have, $$ \begin{aligned} \frac{1}{r^2} & +\frac{1}{r_1^2}+\frac{1}{r_2^2}+\frac{1}{r_3^2} \\ & =\frac{s^2}{\Delta^2}+\frac{(s-a)^2}{\Delta^2}+\frac{(s-b)^2}{\Delta^2}+\frac{(s-c)^2}{\Delta^2} \\ & =\frac{s^2+(s-a)^2+(s-b)^2+(s-c)^2}{\Delta^2} \\ & =\frac{4 s^2-2 s(a+b+c)+a^2+b^2+c^2}{\Delta^2} \\ & =\frac{4 s^2-2 s(2 s)+a^2+b^2+c^2}{\Delta^2} \\ & =\frac{4 s^2-4 s^2+a^2+b^2+c^2}{\Delta^2}=\frac{a^2+b^2+c^2}{\Delta^2} \end{aligned} $$

Hence, the correct answer is option: 'B'.

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