Question

If $\frac{a}{\sqrt{bc}}-2=\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}},$ where $a,b,c>0$, then family of lines $\sqrt{a}x+\sqrt{b}y+\sqrt{c}=0$ passes though the fixed point given by

• A. $(1,1)$
• B. $(1,-2)$
• C. $(-1,2)$
• D. $(1,-1)$

Answer 1

The correct option is D $(1,-1)$

$\frac{a}{\sqrt{bc}}-2=\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}$
$\Rightarrow a=b+c+2\sqrt{bc}$
$\Rightarrow a=(\sqrt{b}+\sqrt{c}{)}^2$
$\Rightarrow (\sqrt{a}-\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+\sqrt{c})=0$
$\Rightarrow \sqrt{a}-\sqrt{b}-\sqrt{c}=0$
Since $\sqrt{a}+\sqrt{b}+\sqrt{c}\ne 0$ ($\because a,b,c>0)$

Comparing with $\sqrt{a}x+\sqrt{b}y+\sqrt{c}=0$, we have $x=1,y=-1$.