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In a $\triangle A B C$, if $r_1=2 r_2=3 r_3$, then $a: b=$
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Verified Answer
The correct answer is:
$5: 4$
Given,
$$
\begin{aligned}
& \quad r_1=2 r_2=3 r_3 \\
& \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c} \\
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{1+2+3}{s-a+s-b+s-c} \\
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{6}{3 s-2 s}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{6}{s} \\
\therefore \quad \frac{1}{s-a} & =\frac{6}{s} \Rightarrow 5 s=6 a \text { and } \frac{2}{s-b}=\frac{6}{s} \Rightarrow 4 s=6 b
\end{aligned}
$$
Hence, $a: b=5 s: 4 s=5: 4$
$$
\begin{aligned}
& \quad r_1=2 r_2=3 r_3 \\
& \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c} \\
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{1+2+3}{s-a+s-b+s-c} \\
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{6}{3 s-2 s}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{1}{s-a}=\frac{2}{s-b}=\frac{3}{s-c}=\frac{6}{s} \\
\therefore \quad \frac{1}{s-a} & =\frac{6}{s} \Rightarrow 5 s=6 a \text { and } \frac{2}{s-b}=\frac{6}{s} \Rightarrow 4 s=6 b
\end{aligned}
$$
Hence, $a: b=5 s: 4 s=5: 4$
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