Find the joint equation of the pair of lines passing through the point (2 , −3) and parallel
to the lines 2𝑥
2 + 3𝑥𝑦 − 9𝑦
2 = 0

To find the joint equation of the pair of lines passing through the point \((2, -3)\) and parallel to the lines given by the equation \(2x^2 + 3xy - 9y^2 = 0\), follow these steps:

### Step 1: Identify the directions of the lines
The given equation \(2x^2 + 3xy - 9y^2 = 0\) represents a pair of lines. We can factor this equation to identify the direction ratios of these lines.

### Step 2: Factor the given equation
The equation can be factored as:
\[
2x^2 + 3xy - 9y^2 = (2x - 3y)(x + 3y) = 0
\]
This shows that the lines are:
\[
2x - 3y = 0 \quad \text{and} \quad x + 3y = 0
\]
These lines are passing through the origin and have direction ratios \((2, 3)\) and \((1, -3)\), respectively.

### Step 3: Form the equations of the new lines
We need to find the equations of lines passing through the point \((2, -3)\) and parallel to these lines.

For the first line \(2x - 3y = 0\):
\[
\text{Equation parallel to } 2x - 3y = 0 \text{ and passing through } (2, -3): 2(x - 2) - 3(y + 3) = 0
\]
\[
2x - 3y - 4 - 9 = 0 \Rightarrow 2x - 3y - 13 = 0
\]

For the second line \(x + 3y = 0\):
\[
\text{Equation parallel to } x + 3y = 0 \text{ and passing through } (2, -3): (x - 2) + 3(y + 3) = 0
\]
\[
x + 3y - 2 + 9 = 0 \Rightarrow x + 3y + 7 = 0

Learn how to find the joint equation of lines passing through a given point and parallel to lines defined by the equation 2x^2 + 3xy - 9y^2 = 0. This problem involves concepts from geometry and equations of lines, suitable for advanced high school students.

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