Question

Separate equations of lines, for a pair of lines, whose equation is $x^2+xy-12y^2=0$ are

• A. $x+4y=0$ and $x+3y=0$
• B. $2x-3y=0$ and $x-4y=0$
• C. $x-6y=0$ and $x-3y=0$
• D. $x+4y=0$ and $x-3y=0$

Answer 1

The correct option is D

$x+4y=0$ and $x-3y=0$

Explanation for the correct option :

Finding the separate equations of lines:

The given equation of the pair of lines is $x^2+xy-12y^2=0$.

The given equation is in the form $a{x}^2+2hxy+b{y}^2=0$ .

We know that if $a{x}^2+2hxy+b{y}^2=0$ represents a pair of straight lines, then the separate equations for the two lines will be

$p_{1}x+q_{1}y+r_1$ and $p_{2}x+q_{2}y+r_2$, where, $a{x}^2+2hxy+b{y}^2=\left(p_{1}x+q_{1}y+r_1\right)\left(p_{2}x+q_{2}y+r_2\right)$

Now, factorizing the equation $x^2+xy-12y^2$ will yield:

$$\begin{gathered} x^2+xy-12y^2=0 \\ \Rightarrow x^2+4xy-3xy-12y^2=0 \\ \Rightarrow x\left(x+4y\right)-3y(x+4y)=0 \\ \Rightarrow \left(x-3y\right)\left(x+4y\right)=0 \end{gathered}$$

Thus, the separate equation of lines are $x+4y=0$ and $x-3y=0$.

Hence, option (D) is the correct answer.