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 through the fixed point given by

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

Answer 1

Answer: (D)

$(-1,1)$

If $\frac{a}{\sqrt{bc}}-2=\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}},$.............
$\frac{a}{\sqrt{bc}}-2=\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\Rightarrow a=b+c+2\sqrt{bc}$
$\Rightarrow a={(\sqrt{b}+\sqrt{c})}^2\Rightarrow (\sqrt{a}-\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+\sqrt{c})=0$
$\Rightarrow =\sqrt{a}-\sqrt{b}-\sqrt{c}=0$(possible) and $\sqrt{a}+\sqrt{b}+\sqrt{c}\ne 0\because (a,b,c>0)$
Comparing with $\sqrt{a}x+\sqrt{b}y+\sqrt{c}=0$
Hence $x=-1,y=1$