Find the point of inersection of lines represented by `2x^2-7xy-4y^2-x+22y-10=0`

Answer

To find the point of intersection of the lines represented by the equation \(2x^2 - 7xy - 4y^2 - x + 22y - 10 = 0\), we will follow these steps:

### Step 1: Partial Differentiation
We need to partially differentiate the given equation with respect to \(x\) and \(y\).

1. **Partial differentiation with respect to \(x\)** (keeping \(y\) constant):
\[ \frac{\partial \phi}{\partial x} = 4x - 7y - 1 = 0 \]

2. **Partial differentiation with respect to \(y\)** (keeping \(x\) constant):
\[ \frac{\partial \phi}{\partial y} = -7x - 8y + 22 = 0 \]

### Step 2: Set Up the System of Equations
From the partial differentiation, we have two equations:
1. \(4x - 7y - 1 = 0\) (Equation 1)
2. \(-7x - 8y + 22 = 0\) (Equation 2)

### Step 3: Solve the System of Equations
We will solve these equations using the elimination method.

1. **Multiply Equation 1 by 7**:
\[ 28x - 49y - 7 = 0 \quad \text{(Equation 3)} \]

2. **Multiply Equation 2 by 4**:
\[ -28x - 32y + 88 = 0 \quad \text{(Equation 4)} \]

### Step 4: Add the Equations
Now, we will add Equation 3 and Equation 4:
\[ (28x - 49y - 7) + (-28x - 32y + 88) = 0 \]

This simplifies to:
\[ -81y + 81 = 0 \]
\[ 81y = 81 \]
\[ y = 1 \]

### Step 5: Substitute Back to Find \(x\)
Now that we have \(y\), we will substitute \(y = 1\) back into Equation 1 to find \(x\):
\[ 4x - 7(1) - 1 = 0 \]
\[ 4x - 7 - 1 = 0 \]
\[ 4x - 8 = 0 \]
\[ 4x = 8 \]
\[ x = 2 \]

### Step 6: Conclusion
The point of intersection of the lines is:
\[ \boxed{(2, 1)} \]