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

Options

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

Answer

To separate the equations of the lines for the given pair of lines represented by the equation \( x^2 + xy - 12y^2 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ x^2 + xy - 12y^2 = 0 \]

### Step 2: Factor the equation
To factor the quadratic equation in terms of \( x \), we can treat it as a standard quadratic equation \( ax^2 + bx + c = 0 \) where:
- \( a = 1 \)
- \( b = y \)
- \( c = -12y^2 \)

### Step 3: Use the quadratic formula
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ x = \frac{-y \pm \sqrt{y^2 - 4(1)(-12y^2)}}{2(1)} \]
This simplifies to:
\[ x = \frac{-y \pm \sqrt{y^2 + 48y^2}}{2} \]
\[ x = \frac{-y \pm \sqrt{49y^2}}{2} \]
\[ x = \frac{-y \pm 7y}{2} \]

### Step 4: Solve for \( x \)
This gives us two cases:
1. \( x = \frac{-y + 7y}{2} = \frac{6y}{2} = 3y \)
2. \( x = \frac{-y - 7y}{2} = \frac{-8y}{2} = -4y \)

### Step 5: Write the equations of the lines
From the above results, we can express the equations of the lines as:
1. \( x - 3y = 0 \) or \( 3y - x = 0 \)
2. \( x + 4y = 0 \) or \( 4y + x = 0 \)

### Final Answer
Thus, the separate equations of the lines are:
1. \( 3y - x = 0 \) (or \( x = 3y \))
2. \( 4y + x = 0 \) (or \( x = -4y \))