The point of intersection of the two lines given by `2x^2-5xy +2y^2-3x+3y+1=0` is

Options

A. A. (-2,2)
B. B. (-3,3)
C. C. (3,3)
D. D. (2,2)

Answer

To find the point of intersection of the two lines given by the equation \(2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0\), we will use partial differentiation. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the equation**: We start with the equation:
\[ 2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0 \]

2. **Partial differentiation with respect to \(x\)**: We differentiate the equation with respect to \(x\) while treating \(y\) as a constant:
\[ \frac{\partial \phi}{\partial x} = 4x - 5y - 3 = 0 \]
This simplifies to:
\[ 4x - 5y = 3 \quad \text{(Equation 1)} \]

3. **Partial differentiation with respect to \(y\)**: Next, we differentiate the equation with respect to \(y\) while treating \(x\) as a constant:
\[ \frac{\partial \phi}{\partial y} = -5x + 4y + 3 = 0 \]
This simplifies to:
\[ 5x - 4y = -3 \quad \text{(Equation 2)} \]

4. **Solve the system of equations**: We now have two equations:
- \(4x - 5y = 3\) (Equation 1)
- \(5x - 4y = -3\) (Equation 2)

We can solve these equations simultaneously. First, we can multiply Equation 1 by 5 and Equation 2 by 4 to eliminate \(x\):
\[ 20x - 25y = 15 \quad \text{(Equation 3)} \]
\[ 20x - 16y = -12 \quad \text{(Equation 4)} \]

5. **Subtract Equation 4 from Equation 3**:
\[ (20x - 25y) - (20x - 16y) = 15 - (-12) \]
This simplifies to:
\[ -25y + 16y = 15 + 12 \]
\[ -9y = 27 \]
Thus, we find:
\[ y = -3 \]

6. **Substitute \(y\) back to find \(x\)**: We can substitute \(y = -3\) into Equation 1:
\[ 4x - 5(-3) = 3 \]
This simplifies to:
\[ 4x + 15 = 3 \]
\[ 4x = 3 - 15 \]
\[ 4x = -12 \]
Thus, we find:
\[ x = -3 \]

7. **Point of intersection**: The point of intersection of the two lines is:
\[ (x, y) = (-3, -3) \]

### Final Answer:
The point of intersection of the two lines is \((-3, -3)\).