Question

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

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

Answer 1

The correct option is D

$(-1,1)$

$\frac{a}{\sqrt{bc}}-2=\sqrt{\frac{b}{c}}+\sqrt{\frac{c}{b}}\dots \left(1\right)$

$a,b,c>0$

$\sqrt{a}x+\sqrt{b}y+\sqrt{c}=0\dots \left(2\right)$

By (1)

$\Rightarrow a=2\sqrt{bc}+b+c$

$\Rightarrow a=(\sqrt{b}+\sqrt{c}{)}^2$

$\Rightarrow (\sqrt{a}-\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}+\sqrt{c})=0$

$\Rightarrow \sqrt{a}-\sqrt{b}-\sqrt{c}=0\dots \left(3\right)$

since $\{\begin{array}{c}\sqrt{a}+\sqrt{b}+\sqrt{c}\ne 0\\ \because a,b,c>0\end{array}$

Comapre (2) & (3)

$\Rightarrow x=-1,y=1$

So, the fixed point is $(-1,1)$