Find the equations to the pair of lines through the origin which are perpendicular represented by `6x^(2) - xy - 12y^(2) = 0`.

Answer

To solve the problem of finding the equations of the pair of lines through the origin which are perpendicular and represented by the equation \(6x^2 - xy - 12y^2 = 0\), we can follow these steps:

### Step 1: Factor the given equation
We start with the equation:
\[ 6x^2 - xy - 12y^2 = 0 \]
We need to factor this quadratic equation in terms of \(x\) and \(y\).

### Step 2: Rearranging the equation
We can rearrange the equation as follows:
\[ 6x^2 - 9xy + 8xy - 12y^2 = 0 \]
This rearrangement helps in grouping the terms for factoring.

### Step 3: Grouping terms
Now we group the terms:
\[ (6x^2 - 9xy) + (8xy - 12y^2) = 0 \]

### Step 4: Factoring by grouping
Next, we factor out common terms from each group:
\[ 3x(2x - 3y) + 4y(2x - 3y) = 0 \]
This can be factored further:
\[ (2x - 3y)(3x + 4y) = 0 \]

### Step 5: Finding the equations of the lines
From the factored form, we can derive the equations of the lines:
1. \(2x - 3y = 0\)
2. \(3x + 4y = 0\)

### Step 6: Finding the slopes of the lines
To find the slopes of these lines, we can rewrite them in slope-intercept form \(y = mx + c\):
- For \(2x - 3y = 0\):
\[ 3y = 2x \implies y = \frac{2}{3}x \quad \text{(slope } m_1 = \frac{2}{3}\text{)} \]
- For \(3x + 4y = 0\):
\[ 4y = -3x \implies y = -\frac{3}{4}x \quad \text{(slope } m_2 = -\frac{3}{4}\text{)} \]

### Step 7: Finding the slopes of the perpendicular lines
The slopes of the perpendicular lines can be found using the relation \(m_1 \cdot m_1' = -1\) and \(m_2 \cdot m_2' = -1\):
- For \(m_1 = \frac{2}{3}\):
\[ m_1' = -\frac{3}{2} \]
- For \(m_2 = -\frac{3}{4}\):
\[ m_2' = \frac{4}{3} \]

### Step 8: Writing the equations of the perpendicular lines
Now we can write the equations of the lines with the new slopes:
1. For \(m_1' = -\frac{3}{2}\):
\[ y = -\frac{3}{2}x \implies 2y + 3x = 0 \]
2. For \(m_2' = \frac{4}{3}\):
\[ y = \frac{4}{3}x \implies 3y - 4x = 0 \]

### Step 9: Final equation of the pair of lines
Now we multiply the two equations to find the equation representing the pair of lines:
\[ (2y + 3x)(3y - 4x) = 0 \]
Expanding this gives:
\[ 6y^2 - 8xy + 6xy - 12x^2 = 0 \implies 6y^2 - 2xy - 12x^2 = 0 \]

### Final Result
Thus, the equations of the pair of lines through the origin which are perpendicular to the given equation are represented by:
\[ 6y^2 - 2xy - 12x^2 = 0 \]