A) \[\frac{a}{b}=\frac{4}{5}\]
B) \[\frac{a}{b}=\frac{5}{4}\]
C) \[a+b-2c=0\]
D) \[2a=b+c\]
Correct Answer: B
Solution :
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\[s-b=2(s-a)\] \[\Rightarrow \] \[\frac{a+b+c}{2}-b=2\left( \frac{a+b+c}{2}-a \right)\] \[\left( \because \,\,s=\frac{a+b+c}{2} \right)\] \[\Rightarrow \] \[a+c-b=2(-a+b+c)\] \[\Rightarrow \] \[3a-c-3b=0\] \[\Rightarrow \] \[3a=3b+c\] ? (i) Now, taking\[(s-c)=3(s-a)\] \[\Rightarrow \] \[\frac{a+b+c}{2}-c=3\left( \frac{a+b+c}{2}-a \right)\] \[\Rightarrow \] \[a+b-c=3(-a+b+c)\] \[\Rightarrow \] \[4a-2b-4c=0\] \[\Rightarrow \] \[4a=2b+4c\] From Eq. (i) and (ii), we get \[6a=6b+2a-b\] \[\Rightarrow \] \[4a=5b\] \[\Rightarrow \] \[\frac{a}{b}=\frac{5}{4}\]
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