the lines `x + y + 1 = 0; 4x + 3y + 4 = 0` and `x + alphay + beta= 0,` where `alpha^2 + beta^2 = 2,` are concurrent

Options

A. ` alpha = 1 , beta = -1 `
B. ` alpha = 1 , beta = pm 1 `
C. ` alpha = -1 , beta = pm 1 `
D. ` alpha = pm 1 , beta = 1 `

Answer

To solve the problem, we need to determine the values of \( \alpha \) and \( \beta \) such that the lines \( x + y + 1 = 0 \), \( 4x + 3y + 4 = 0 \), and \( x + \alpha y + \beta = 0 \) are concurrent. This means that the three lines intersect at a single point, which can be determined using the determinant of their coefficients.

### Step-by-Step Solution:

1. **Write the equations of the lines:**
- Line 1: \( x + y + 1 = 0 \) can be rewritten as \( 1x + 1y + 1 = 0 \).
- Line 2: \( 4x + 3y + 4 = 0 \) can be rewritten as \( 4x + 3y + 4 = 0 \).
- Line 3: \( x + \alpha y + \beta = 0 \) can be rewritten as \( 1x + \alpha y + \beta = 0 \).

2. **Set up the determinant for concurrency:**
The lines are concurrent if the determinant of the coefficients is equal to zero:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 4 & 3 & 4 \\ 1 & \alpha & \beta \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
Expanding the determinant:
\[ = 1 \begin{vmatrix} 3 & 4 \\ \alpha & \beta \end{vmatrix} - 1 \begin{vmatrix} 4 & 4 \\ 1 & \beta \end{vmatrix} + 1 \begin{vmatrix} 4 & 3 \\ 1 & \alpha \end{vmatrix} \]
Calculating each of the 2x2 determinants:
- First determinant: \( 3\beta - 4\alpha \)
- Second determinant: \( 4\beta - 4 \)
- Third determinant: \( 4\alpha - 3 \)

Thus, we have:
\[ 1(3\beta - 4\alpha) - 1(4\beta - 4) + 1(4\alpha - 3) = 0 \]
Simplifying this gives:
\[ 3\beta - 4\alpha - 4\beta + 4 + 4\alpha - 3 = 0 \]
Combining like terms:
\[ -\beta + 1 = 0 \]
Therefore:
\[ \beta = 1 \]

4. **Use the condition \( \alpha^2 + \beta^2 = 2 \):**
Now substitute \( \beta = 1 \) into the equation:
\[ \alpha^2 + 1^2 = 2 \]
This simplifies to:
\[ \alpha^2 + 1 = 2 \implies \alpha^2 = 1 \]
Thus, we find:
\[ \alpha = \pm 1 \]

### Final Result:
The values of \( \alpha \) and \( \beta \) are:
- \( \beta = 1 \)
- \( \alpha = \pm 1 \)