If `xy+x+y+1=0,x+ay-3=0` are concurrent then a=

Options

A. 3
B. `4`
C. 2
D. none of these

Answer

To determine the value of \( a \) for which the lines represented by the equations \( xy + x + y + 1 = 0 \) and \( x + ay - 3 = 0 \) are concurrent, we can follow these steps:

### Step 1: Factor the first equation
The first equation is given as:
\[ xy + x + y + 1 = 0 \]
We can rearrange this equation:
\[ xy + x + y + 1 = (x + 1)(y + 1) = 0 \]
This gives us two lines:
1. \( x + 1 = 0 \) (or \( x = -1 \))
2. \( y + 1 = 0 \) (or \( y = -1 \))

### Step 2: Write the second equation
The second equation is:
\[ x + ay - 3 = 0 \]
This can be rewritten in standard form as:
\[ x + ay = 3 \]

### Step 3: Set up the determinant for concurrency
For the lines to be concurrent, the determinant of the coefficients must equal zero. We can set up the determinant using the coefficients of the lines:
\[ \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & a & -3 \end{vmatrix} = 0 \]

### Step 4: Calculate the determinant
Now, we calculate the determinant:
\[ \begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & a & -3 \end{vmatrix} = 1 \cdot (1 \cdot (-3) - 1 \cdot a) - 0 + 1 \cdot (0 \cdot a - 1 \cdot 1) \]
This simplifies to:
\[ 1 \cdot (-3 - a) + 0 - 1 = -3 - a - 1 = -4 - a \]

### Step 5: Set the determinant to zero
Setting the determinant equal to zero for concurrency:
\[ -4 - a = 0 \]

### Step 6: Solve for \( a \)
Now, solving for \( a \):
\[ a = -4 \]

### Conclusion
Thus, the value of \( a \) for which the lines are concurrent is:
\[ \boxed{-4} \]