Q66 / 160 · AP EAMCET 2023 (16 May Shift 2) · Mathematics

In $\triangle \mathrm{ABC}$, if $\angle \mathrm{A}=90^{\circ}$, then $\left(\mathrm{r}_2-\mathrm{r}_1\right)\left(\mathrm{r}_3-\mathrm{r}_1\right)=$

  • A
    $\mathrm{r}_2 \mathrm{r}_3$
  • Correct answer
    B
    $2 r_2 r_3$
  • C
    $4 r_2 r_3$
  • D
    $2$

Solution

In $\triangle \mathrm{ABC} \angle \mathrm{A}=90^{\circ}$ $\begin{aligned} & \Rightarrow b^2+c^2=a^2 \\ & \Rightarrow \frac{1}{2}(b+c-a)^2=(b-a)(c-a) \\ & \Rightarrow(b-a)(c-a)=2(s-a)^2 \\ & \Rightarrow \frac{(b-a)(c-a)}{(s-a)^2}=2 \end{aligned}$ $\begin{aligned} & \Rightarrow\left(1-\frac{s-b}{s-a}\right)\left(1-\frac{s-c}{s-a}\right)=2 \\ & \Rightarrow\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-b}}\right)\left(1-\frac{\frac{\Delta}{s-a}}{\frac{\Delta}{s-c}}\right)=2 \\ & \Rightarrow\left(r_2-r_1\right)\left(r_3-r_1\right)=2 r_1 r_3\end{aligned}$

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

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