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 $r_1=2r_2=3r_3$
$\Rightarrow \frac{\Delta }{s-a}=\frac{2\Delta }{(s-b)}=\frac{3\Delta }{(s-c)}$
$\Rightarrow s-b=2(s-a)$ and $(s-c)=3(s-a)$
Taking $s-b=2(s-a)$
$\Rightarrow \frac{a+b+c}{2}-b=2\left(\frac{a+b+c}{2}-a\right)$
$\left[\because s=\frac{a+b+c}{2}\right]$
$\Rightarrow a+c-b=2(-a+b+c)$
$\Rightarrow 3a-c-3b=0$
$\Rightarrow 3a-3b+c\dots (i)$
Now, taking $(s-c)=3(s-a)$
$\Rightarrow \frac{a+b+c}{2}-c=3\left(\frac{a+b+c}{2}-a\right)$
$\Rightarrow a+b-c=3(-a+b+c)$
$\Rightarrow 4a-2b-4c=0$
$\Rightarrow 4a=2b+4c$
$\Rightarrow 2a=b+2c\dots (ii)$
From Eqs. (i) and (ii), we get
$6a=6b+2a-b$
$\Rightarrow 4a=5b$
$\Rightarrow \frac{a}{b}=\frac{5}{4}$