Q68 / 160 · AP EAMCET 2024 (20 May Shift 2) · Mathematics

In $\triangle A B C, b c-r_2 r_3=$

  • Correct answer
    A
    $r r_1$
  • B
    $r_2$
  • C
    $r_1$
  • D
    $a r_1$

Solution

$\because r r_1+r_2 r_3=\frac{\Delta}{s} \times \frac{\Delta}{s-a}+\frac{\Delta}{s-b} \times \frac{\Delta}{s-c}$ $\begin{aligned} & =\frac{\Delta^2[(s-b)(s-c)+s(s-a)]}{s(s-a)(s-b)(s-c)} \\ & =(s-b)(s-c)+s(s-a) \\ & \qquad[\because \Delta=\sqrt{s(s-a)(s-b)(s-c)}] \\ & =s^2-s(b+c)+b c+s^2-s a \\ & =2 s^2-s(a+b+c)+b c \qquad [\because 2 s=a+b+c] \\ & =2 s^2-2 s^2+b c \\ & \therefore r r_1+r_2 r_3=b c \Rightarrow r r_1=b c-r_2 r_3\end{aligned}$

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

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