In a $△ A B C$ , if $r_{1} = 2 r_{2} = 3 r_{3}$ , then

\frac{a}{b} = \frac{4}{5}

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\frac{a}{b} = \frac{5}{4}

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a + b - 2 c = 0

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2 a = b + c

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Solution

The correct option is A $\frac{a}{b} = \frac{5}{4}$
Given, $r_{1} = 2 r_{2} = 3 r_{3}$
$∴ \frac{△}{s - a} = \frac{2 △}{s - b} = \frac{3 △}{s - c}$
$\Rightarrow (s - b) = 2 (s - a)$ and $(s - c) = 3 (s - a)$
$\Rightarrow (\frac{a + b + c}{2} - b) = 2 (\frac{a + b + c}{2} - a)$
and $(\frac{a + b + c}{2} - c) = 3 (\frac{a + b + c}{2} - a)$
$\Rightarrow a + c - b = 2 (- a + b + c)$
and $a + b - c = 3 (- a + b + c)$
$\Rightarrow 3 a = 3 b + c$ ...... $(i)$
and $2 a = b + 2 c$ ....... $(i i)$
Multiply $(i)$ by $2$ , then subtract $(i i)$ from $(i)$
$\Rightarrow 4 a = 5 b$
$\Rightarrow \frac{a}{b} = \frac{5}{4}$

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