Question

If $\dfrac{a}{{\sqrt {bc} }} - 2 = \sqrt {\dfrac{b}{c}} + \sqrt {\dfrac{c}{b}}$ , 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. $\left( {1,1} \right)$
B. $\left( {1, - 2} \right)$
C. $\left( { - 1,2} \right)$
D. $\left( { - 1,1} \right)$

Answer 1

Hint : In order to get the fixed point, solve the given equation $\dfrac{a}{{\sqrt {bc} }} - 2 = \sqrt {\dfrac{b}{c}} + \sqrt {\dfrac{c}{b}}$ , obtain an equation in terms of the variables $a,b,c$ . Then compare the equation with $\sqrt a x + \sqrt b y + \sqrt c = 0$ , find the value of $x,y$ and place it in the fixed point $\left( {x,y} \right)$ .

Complete step by step solution:
We are given with an equation $\dfrac{a}{{\sqrt {bc} }} - 2 = \sqrt {\dfrac{b}{c}} + \sqrt {\dfrac{c}{b}}$ .
Solving this equation, step by step:
For that multiplying and dividing $- 2$ by $\sqrt {bc}$ , in order to have a common denominator on the left side, and we get:
$\dfrac{a}{{\sqrt {bc} }} - 2\dfrac{{\sqrt {bc} }}{{\sqrt {bc} }} = \sqrt {\dfrac{b}{c}} + \sqrt {\dfrac{c}{b}} \\\ \Rightarrow \dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} = \sqrt {\dfrac{b}{c}} + \sqrt {\dfrac{c}{b}} \;$
Since, we know that roots can be written as $\sqrt {\dfrac{b}{c}} = \dfrac{{\sqrt b }}{{\sqrt c }}$ and $\sqrt {bc} = \sqrt b \sqrt c$ .
So, using this writing the right-side equation as:
$\dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} = \dfrac{{\sqrt b }}{{\sqrt c }} + \dfrac{{\sqrt c }}{{\sqrt b }}$
Multiplying and dividing the first operand on the right side by $\sqrt b$ , and multiplying and dividing the second operand on the right side by $\sqrt c$ , in order to get a common denominator:
Applying this, we get:
$\dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} = \dfrac{{\sqrt b }}{{\sqrt c }} \times \dfrac{{\sqrt b }}{{\sqrt b }} + \dfrac{{\sqrt c }}{{\sqrt b }} \times \dfrac{{\sqrt c }}{{\sqrt c }}$
Solving the right-hand side:
Since, we know that $\sqrt x .\sqrt x = x$ , that implies:
$\dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} = \dfrac{b}{{\sqrt {bc} }} + \dfrac{c}{{\sqrt {bc} }} \\\ \Rightarrow \dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} = \dfrac{{b + c}}{{\sqrt {bc} }} \;$
Multiplying both the sides by $\sqrt {bc}$ :
$\dfrac{{a - 2\sqrt {bc} }}{{\sqrt {bc} }} \times \sqrt {bc} = \dfrac{{b + c}}{{\sqrt {bc} }} \times \sqrt {bc} \\\ \Rightarrow a - 2\sqrt {bc} = b + c \;$
Adding both the sides by $2\sqrt {bc}$ :
$\Rightarrow a - 2\sqrt {bc} = b + c \\\ \Rightarrow a - 2\sqrt {bc} + 2\sqrt {bc} = b + c + 2\sqrt {bc} \\\ \Rightarrow a = b + c + 2\sqrt {bc} \;$
Since, we know that $\sqrt x .\sqrt x = x$ , so $b, c$ and $a$ can be written as $b = {\left( {\sqrt b } \right)^2}$ , $a = {\left( {\sqrt a } \right)^2}$ , $c = {\left( {\sqrt c } \right)^2}$ .
Substituting these values in the above equation, we get:
$\Rightarrow a = b + c + 2\sqrt {bc} \\\ \Rightarrow {\left( {\sqrt a } \right)^2} = {\left( {\sqrt b } \right)^2} + {\left( {\sqrt c } \right)^2} + 2\sqrt {bc} \;$
On the right-hand side, we can see that it perfectly suits the formula: ${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$ .
So, replacing ${\left( {\sqrt b } \right)^2} + {\left( {\sqrt c } \right)^2} + 2\sqrt {bc}$ by ${\left( {\sqrt b + \sqrt c } \right)^2}$ , we write it as:
${\left( {\sqrt a } \right)^2} = {\left( {\sqrt b + \sqrt c } \right)^2}$
Subtracting both the sides by ${\left( {\sqrt b + \sqrt c } \right)^2}$ , we get:
${\left( {\sqrt a } \right)^2} - {\left( {\sqrt b + \sqrt c } \right)^2} = {\left( {\sqrt b + \sqrt c } \right)^2} - {\left( {\sqrt b + \sqrt c } \right)^2} \\\ \Rightarrow {\left( {\sqrt a } \right)^2} - {\left( {\sqrt b + \sqrt c } \right)^2} = 0 \;$
Since, we know that ${x^2} - {y^2} = \left( {x + y} \right)\left( {x - y} \right)$ , so applying this in the above equation:
${\left( {\sqrt a } \right)^2} - {\left( {\sqrt b + \sqrt c } \right)^2} = 0 \\\ \Rightarrow \left( {\sqrt a + \left( {\sqrt b + \sqrt c } \right)} \right)\left( {\sqrt a - \left( {\sqrt b + \sqrt c } \right)} \right) = 0 \\\$
Opening the inner parenthesis:
$\left( {\sqrt a + \left( {\sqrt b + \sqrt c } \right)} \right)\left( {\sqrt a - \left( {\sqrt b + \sqrt c } \right)} \right) = 0 \\\ \Rightarrow \left( {\sqrt a + \sqrt b + \sqrt c } \right)\left( {\sqrt a - \sqrt b - \sqrt c } \right) = 0 \;$
So, from the above equation, we can see that either $\left( {\sqrt a + \sqrt b + \sqrt c } \right) = 0$ or $\left( {\sqrt a - \sqrt b - \sqrt c } \right) = 0$ .
But it’s given that $a,b,c > 0$ , that implies $\left( {\sqrt a + \sqrt b + \sqrt c } \right) \ne 0$ as the square root values will give a positive number and when added, the terms would never result into zero.
That gives $\left( {\sqrt a - \sqrt b - \sqrt c } \right) = 0$ , taking $- 1$ common from the equation and which can be written as
$\Rightarrow \left( {\sqrt a - \sqrt b - \sqrt c } \right) = 0 \\\ \Rightarrow - 1\left( { - \sqrt a + \sqrt b + \sqrt c } \right) = 0 \;$ .
Dividing both sides by $- 1$ , we get:
$\Rightarrow \left( { - 1} \right)\sqrt a + \sqrt b \left( 1 \right) + \sqrt c = 0$
Since, it’s given that the family of lines passes through the fixed point, so the equations would be the same.
So, comparing $\left( { - 1} \right)\sqrt a + \sqrt b \left( 1 \right) + \sqrt c = 0$ with another equation given $\sqrt a x + \sqrt b y + \sqrt c = 0$ , we get:
$x = - 1 \\\ y = 1 \;$
Therefore, the fixed point becomes $\left( {x,y} \right) = \left( { - 1,1} \right)$ , which matches with the option D.
Hence, Option D is correct.
So, the correct answer is “Option D”.

Note : It’s always preferred to solve step by step for ease, rather than solving at once, otherwise it would lead to confusion and there is a huge chance of error.
Writing of roots from one form to another is mandatory, like $\sqrt {\dfrac{b}{c}} = \dfrac{{\sqrt b }}{{\sqrt c }}$ and $\sqrt {bc} = \sqrt b \sqrt c$.