Question

In a $ \Delta\,ABC $ , if $ r_{1}=2r_{2}=3r_{3} $ , then

• A. $ \frac{a}{b}=\frac{4}{5} $
• B. $ \frac{a}{b}=\frac{5}{4} $
• C. $ a + b − 2c = 0 $
• D. $ 2a = b + c $

Answer 1

Answer: (B)

We have $r_{1}=2 r_{2}=3 r_{3}$
$\Rightarrow \, \frac{\Delta}{s-a}=\frac{2 \Delta}{(s-b)}=\frac{3 \Delta}{(s-c)}$
$\Rightarrow \,s-b=2(s-a)$ and $(s-c)=3(s-a)$
Taking $s-b=2(s-a)$
$\Rightarrow \,\frac{a+b+c}{2}-b=2\left(\frac{a+b+c}{2}-a\right)$
$\left[\because s=\frac{a+b+c}{2}\right]$
$\Rightarrow \, a+c-b=2(-a+b+c)$
$\Rightarrow \, 3 a-c-3 b=0$
$\Rightarrow \, 3 a-3 b+c\,\,\,\,\,\,\dots(i)$
Now, taking $(s-c)=3(s-a)$
$\Rightarrow \, \frac{a+b+c}{2}-c=3\left(\frac{a+b+c}{2}-a\right)$
$\Rightarrow \, a+b-c=3(-a+b+c)$
$\Rightarrow \,4 a-2 b-4 c=0$
$\Rightarrow \, 4 a=2 b+4 c$
$\Rightarrow \, 2 a=b+2 c\,\,\,\,\,\dots(ii)$
From Eqs. (i) and (ii), we get
$6 a=6 b+2 a-b$
$\Rightarrow \, 4 a=5 b$
$\Rightarrow \, \frac{a}{b}=\frac{5}{4}$