If the lines `3x + 4y + 1 = 0`, `5x+lambday+3=0` and `2x+y-1=0` are concurrent, then `lambda` is equal to-

Options

A. -8
B. 8
C. 4
D. -4

Answer

To find the value of \( \lambda \) such that the lines \( 3x + 4y + 1 = 0 \), \( 5x + \lambda y + 3 = 0 \), and \( 2x + y - 1 = 0 \) are concurrent, we can use the determinant condition for concurrency of three lines.

### Step-by-Step Solution:

1. **Write the equations in standard form**:
The equations of the lines are:
- Line 1: \( 3x + 4y + 1 = 0 \) (Here, \( a_1 = 3, b_1 = 4, c_1 = 1 \))
- Line 2: \( 5x + \lambda y + 3 = 0 \) (Here, \( a_2 = 5, b_2 = \lambda, c_2 = 3 \))
- Line 3: \( 2x + y - 1 = 0 \) (Here, \( a_3 = 2, b_3 = 1, c_3 = -1 \))

2. **Set up the determinant**:
The lines are concurrent if the following determinant is equal to zero:
\[ \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = 0 \]
Substituting the values, we have:
\[ \begin{vmatrix} 3 & 4 & 1 \\ 5 & \lambda & 3 \\ 2 & 1 & -1 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
The determinant can be calculated as follows:
\[ = 3 \begin{vmatrix} \lambda & 3 \\ 1 & -1 \end{vmatrix} - 4 \begin{vmatrix} 5 & 3 \\ 2 & -1 \end{vmatrix} + 1 \begin{vmatrix} 5 & \lambda \\ 2 & 1 \end{vmatrix} \]
Now, we compute each of the 2x2 determinants:
- \( \begin{vmatrix} \lambda & 3 \\ 1 & -1 \end{vmatrix} = \lambda(-1) - 3(1) = -\lambda - 3 \)
- \( \begin{vmatrix} 5 & 3 \\ 2 & -1 \end{vmatrix} = 5(-1) - 3(2) = -5 - 6 = -11 \)
- \( \begin{vmatrix} 5 & \lambda \\ 2 & 1 \end{vmatrix} = 5(1) - \lambda(2) = 5 - 2\lambda \)

Substituting these back into the determinant:
\[ = 3(-\lambda - 3) - 4(-11) + (5 - 2\lambda) \]
Simplifying this:
\[ = -3\lambda - 9 + 44 + 5 - 2\lambda \]
\[ = -5\lambda + 40 \]

4. **Set the determinant to zero**:
To find \( \lambda \):
\[ -5\lambda + 40 = 0 \]
\[ 5\lambda = 40 \]
\[ \lambda = 8 \]

### Final Answer:
Thus, the value of \( \lambda \) is \( 8 \).