The point of intersection of the pair of straight lines given by `6x^(2)+5xy-4y^(2)+7x+13y-2=0`, is

Options

A. `(1,1)`
B. `(1,-1)`
C. `(-1,1)`
D. `(-1,-1)`

Answer

To find the point of intersection of the pair of straight lines given by the equation \(6x^2 + 5xy - 4y^2 + 7x + 13y - 2 = 0\), we will follow these steps:

### Step 1: Differentiate the equation with respect to \(x\)
We will partially differentiate the given equation with respect to \(x\):
\[ \frac{\partial}{\partial x}(6x^2 + 5xy - 4y^2 + 7x + 13y - 2) = 0 \]
This gives us:
\[ 12x + 5y + 7 = 0 \]
Let this be Equation (1).

### Step 2: Differentiate the equation with respect to \(y\)
Next, we will partially differentiate the given equation with respect to \(y\):
\[ \frac{\partial}{\partial y}(6x^2 + 5xy - 4y^2 + 7x + 13y - 2) = 0 \]
This gives us:
\[ 5x - 8y + 13 = 0 \]
Let this be Equation (2).

### Step 3: Solve the system of equations
Now we need to solve the two equations obtained from differentiation:
1. \(12x + 5y + 7 = 0\) (Equation 1)
2. \(5x - 8y + 13 = 0\) (Equation 2)

To eliminate \(y\), we can multiply Equation (1) by \(8\) and Equation (2) by \(5\):
- From Equation (1):
\[ 8(12x + 5y + 7) = 0 \implies 96x + 40y + 56 = 0 \quad \text{(Equation 3)} \]
- From Equation (2):
\[ 5(5x - 8y + 13) = 0 \implies 25x - 40y + 65 = 0 \quad \text{(Equation 4)} \]

### Step 4: Add Equations (3) and (4)
Now we add Equation (3) and Equation (4):
\[ (96x + 40y + 56) + (25x - 40y + 65) = 0 \]
This simplifies to:
\[ 121x + 121 = 0 \]
From this, we can solve for \(x\):
\[ 121x = -121 \implies x = -1 \]

### Step 5: Substitute \(x\) back to find \(y\)
Now we substitute \(x = -1\) back into Equation (1) to find \(y\):
\[ 12(-1) + 5y + 7 = 0 \]
This simplifies to:
\[ -12 + 5y + 7 = 0 \implies 5y - 5 = 0 \implies 5y = 5 \implies y = 1 \]

### Conclusion
Thus, the point of intersection of the pair of straight lines is \((-1, 1)\).