Q111 / 170 · Manipal (MET) 2019 · Mathematics

In $\triangle A B C, \frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3}$ is equal to

  • Correct answer
    A
    $0$
  • B
    $a b c$
  • C
    $a+b+c$
  • D
    $a b+b c+c a$

Solution

$\begin{aligned} & \text { Now, } \frac{b-c}{r_1}+\frac{c-a}{r_2}+\frac{a-b}{r_3} \\ & =\frac{[(b-c)](s-a)+(c-a)(s-b)+(a-b)}{\Delta}=0\end{aligned}$

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

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