Question

If $\frac{bx}{c}+\frac{cy}{a}+\frac{az}{b}=\frac{a^2+b^2+c^2}{k}$, then the value of $k$ is

• A. $R$
• B. $S$
• C. $2R$
• D. $\frac{3}{2R}$

Answer 1

Answer: (C)

$2R$

Clearly, $2\Delta =ax=by=cz$

$\therefore \frac{b^{2}xa+c^{2}yb+a^{2}zc}{abc}=\frac{a^2+b^2+c^2}{k}$

$\Rightarrow \frac{b^2\left(2\Delta \right)+c^2\left(2\Delta \right)+a^2\left(2\Delta \right)}{abc}=\frac{a^2+b^2+c^2}{k}$

$\Rightarrow \frac{2\Delta }{abc}=\frac{1}{k}\Rightarrow k=\frac{abc}{2\Delta }=2R[\because \Delta =\frac{abc}{4R}]$