If the pair of straight lines `xy - x - y +1=0` & the line `ax+2y-3=0` are concurrent then `a =`

Options

A. `-1`
B. 0
C. 3
D. 1

Answer

To solve the problem, we need to determine the value of \( a \) such that the lines represented by the equations \( xy - x - y + 1 = 0 \) and \( ax + 2y - 3 = 0 \) are concurrent.

### Step-by-Step Solution:

1. **Rewrite the first equation**:
The equation \( xy - x - y + 1 = 0 \) can be factored. We can rearrange it as follows:
\[ xy - x - y + 1 = 0 \implies (x - 1)(y - 1) = 0 \]
This gives us two lines:
\[ y - 1 = 0 \quad \text{(Line 1: } l_1) \quad \text{and} \quad x - 1 = 0 \quad \text{(Line 2: } l_2) \]

2. **Identify the equations of the lines**:
- Line 1: \( y = 1 \)
- Line 2: \( x = 1 \)
- Line 3: \( ax + 2y - 3 = 0 \)

3. **Express Line 3 in standard form**:
The third line can be rewritten as:
\[ 2y = -ax + 3 \implies y = -\frac{a}{2}x + \frac{3}{2} \]

4. **Set up the determinant condition for concurrency**:
For three lines to be concurrent, the determinant of their coefficients must be zero. The lines can be represented in the form:
\[ a_1x + b_1y + c_1 = 0, \quad a_2x + b_2y + c_2 = 0, \quad a_3x + b_3y + c_3 = 0 \]
Here, we have:
- For \( l_1: 0x + 1y - 1 = 0 \) (coefficients: \( a_1 = 0, b_1 = 1, c_1 = -1 \))
- For \( l_2: 1x + 0y - 1 = 0 \) (coefficients: \( a_2 = 1, b_2 = 0, c_2 = -1 \))
- For \( l_3: ax + 2y - 3 = 0 \) (coefficients: \( a_3 = a, b_3 = 2, c_3 = -3 \))

5. **Set up the determinant**:
The determinant is given by:
\[ \begin{vmatrix} 0 & 1 & -1 \\ 1 & 0 & -1 \\ a & 2 & -3 \end{vmatrix} = 0 \]

6. **Calculate the determinant**:
Expanding the determinant:
\[ = 0 \cdot (0 \cdot -3 - (-1) \cdot 2) - 1 \cdot (1 \cdot -3 - (-1) \cdot a) + (-1) \cdot (1 \cdot 2 - 0 \cdot a) \]
Simplifying:
\[ = 0 - (-3 + a) - 2 = 0 \]
\[ = 3 - a - 2 = 0 \implies 1 - a = 0 \implies a = 1 \]

7. **Final Result**:
Thus, the value of \( a \) for which the lines are concurrent is:
\[ \boxed{1} \]