Joint equation of pair of lines through `(3,-2)` and parallel to `x^2-4xy+3y^2=0` is

Options

A. `x^2+3y^2-4xy-14y+24y+45=0`
B. `x^2+3y^2+4xy-14+2y+45=0`
C. `x^2+3y^2+4xy-14x+24y-45=0`
D. `x^2+3y^2+4xy-14x-24y-45=0`

Explanation

To find the joint equation of the pair of lines through the point (3, -2) and parallel to the lines represented by the equation \(x^2 - 4xy + 3y^2 = 0\), we can follow these steps:

### Step 1: Identify the slopes of the lines represented by the given equation
The given equation \(x^2 - 4xy + 3y^2 = 0\) can be factored to find the slopes of the lines it represents. The equation can be rewritten as:
\[ (x - 3y)(x - y) = 0 \]
This indicates two lines:
1. \(x - 3y = 0\) (or \(y = \frac{1}{3}x\)) with slope \(m_1 = \frac{1}{3}\)
2. \(x - y = 0\) (or \(y = x\)) with slope \(m_2 = 1\)