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 point

• A. $(1,1)$
• B. $(-1,1)$
• C. $(1,3)$
• D. $(1,4)$

Answer 1

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

Given,

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

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

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

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

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

$(\sqrt{a}{)}^2-(\sqrt{b}+\sqrt{c}{)}^2=0$

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

$(\sqrt{a}+\sqrt{b}+\sqrt{c})\ne 0$

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

$\sqrt{a}x+\sqrt{b}y+\sqrt{c}=0$

$x=-1,y=1$

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

$\sqrt{a}-\sqrt{b}-\sqrt{c}=0$

$(x,y)\Rightarrow (-1,1)$