Q66 / 160 · AP EAMCET 2023 (17 May Shift 2) · Mathematics

In $\triangle \mathrm{ABC}, \frac{\mathrm{r}_1-\mathrm{r}}{\mathrm{a}}+\frac{\mathrm{r}_2-\mathrm{r}}{\mathrm{b}}=$

  • A
    $\frac{a}{r_3}$
  • B
    $\frac{\mathrm{b}}{\mathrm{r}_3}$
  • Correct answer
    C
    $\frac{\mathrm{c}}{\mathrm{r}_3}$
  • D
    1

Solution

$\frac{r_1-r}{a}+\frac{r_{2-r}}{b}=\frac{\frac{\Delta}{s-a}-\frac{\Delta}{s}}{a}+\frac{\frac{\Delta}{s-b}-\frac{\Delta}{s}}{b}$ $\begin{aligned} & =\frac{\Delta}{a} \cdot \frac{s-s+a}{s(\mathrm{~s}-a)}+\frac{\Delta}{b} \frac{s-s+b}{s(\mathrm{~s}-\mathrm{b})}=\frac{\Delta}{s(s-a)}+\frac{\Delta}{s(s-b)} \\ & =\frac{\Delta}{s} \cdot \frac{s-b+s-a}{(s-a)(s-b)} \\ & =\frac{\Delta(a+b+c-b-a)(s-c)}{\Delta^2}=\frac{c}{\frac{\Delta}{s-c}}=\frac{c}{r_3}\end{aligned}$

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

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