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 $.