Question

The lines ${a^2}{x^2} + bc{y^2} = a\left( {b + c} \right)xy$ will be coincident, if
A.$a = 0$ or $b = c$
B.$a = b$ or $a = c$
C.$c = 0$ or $a = b$
D.$a = b + c$

Answer 1

Hint : Here in this question, we need to find the condition of if the given equation of line will be incident. For this, first we need to compare the given equation of line with the general homogeneous equation of second degree, then if lines are real and coincident apply a condition ${h^2} - ab = 0$ and on further simplification we get the required solution.

Complete step-by-step answer :
If the index of every variable in an expression is a non-negative integer and the sum of the indices of the variable in each term of an expression is the same ( say $n$), then the expression is called a homogeneous expression of degree ' $n$ ' in these variables. The further equation obtained by equating the above expression with zero is called the homogeneous equation of degree ' $n$ ' in these variables.
The second degree Homogeneous Equation in x and y:
Any equation in the form $a{x^2} + 2hxy + b{y^2} = 0$ is called a second degree homogeneous equation in x and y. where a, h, b are real and not all zero.
Consider the given equation
${a^2}{x^2} + bc{y^2} = a\left( {b + c} \right)xy$
On rearranging, it can be written as
$\Rightarrow \,\,\,\,{a^2}{x^2} + bc{y^2} - a\left( {b + c} \right)xy = 0$ -----(1)
Now, comparing the equation (1) with the general homogeneous second degree equation $a{x^2} + 2hxy + b{y^2} = 0$.
In equation (1) $a = {a^2}$, $b = bc$, $h = - \dfrac{{a\left( {b + c} \right)}}{2}$
In second degree homogeneous equation if the nature of lines are coincident, then
${h^2} - ab = 0$
On substituting the values, we have
$\Rightarrow \,\,\,{\left( { - \dfrac{{a\left( {b + c} \right)}}{2}} \right)^2} - {a^2}bc = 0$
$\Rightarrow \,\,\,\dfrac{{{a^2}{{\left( {b + c} \right)}^2}}}{{{2^2}}} - {a^2}bc = 0$
$\Rightarrow \,\,\,\dfrac{{{a^2}{{\left( {b + c} \right)}^2}}}{4} - {a^2}bc = 0$
Take 4 as LCM in LHS
$\Rightarrow \,\,\,{a^2}{\left( {b + c} \right)^2} - 4{a^2}bc = 0$
Take ${a^2}$ as common
$\Rightarrow \,\,\,{a^2}\left( {{{\left( {b + c} \right)}^2} - 4bc} \right) = 0$
Apply a algebraic formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$, then we have
$\Rightarrow \,\,\,{a^2}\left( {{b^2} + {c^2} + 2bc - 4bc} \right) = 0$
On simplification, we get
$\Rightarrow \,\,\,{a^2}\left( {{b^2} + {c^2} - 2bc} \right) = 0$
By the algebraic formula ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$, then
$\Rightarrow \,\,\,{a^2}{\left( {b - c} \right)^2} = 0$
Equating the each factor with zero
$\Rightarrow \,\,\,{a^2} = 0$ or ${\left( {b - c} \right)^2} = 0$
$\Rightarrow \,\,\,\,a = 0$ or $b - c = 0$
$\therefore \,\,\,\,a = 0$ or $b = c$
Hence, it’s a required solution.
Therefore, Option (1) is the correct answer.
So, the correct answer is “Option 1”.

Note : Remember $a{x^2} + 2hxy + b{y^2} = 0$ is the second degree homogeneous equation, the every second degree homogeneous equation in x and y in general represents a pair of lines through origin. Then the Nature of lines:
If ${h^2} - ab \geqslant 0$, then the lines are real and distinct.
If ${h^2} - ab = 0$, then the lines are real and coincident.
If ${h^2} - ab < 0$, then the lines are not real and can't be drawn.