The lines `ax+2y+1=0, bx+3y+1=0` and `cx+4y+1=0` are concurrent of a,b,c are in

Options

A. A.P
B. G.P
C. H.P
D. None

Answer

To determine the relationship between the coefficients \(a\), \(b\), and \(c\) of the lines \(ax + 2y + 1 = 0\), \(bx + 3y + 1 = 0\), and \(cx + 4y + 1 = 0\) when they are concurrent, we can use the condition that the determinant formed by the coefficients must equal zero.

### Step-by-Step Solution:

1. **Set Up the Determinant**:
The lines are concurrent if the following determinant is equal to zero:
\[ \begin{vmatrix} a & 2 & -1 \\ b & 3 & -1 \\ c & 4 & -1 \end{vmatrix} = 0 \]

2. **Calculate the Determinant**:
Expanding the determinant along the first row:
\[ = a \begin{vmatrix} 3 & -1 \\ 4 & -1 \end{vmatrix} - 2 \begin{vmatrix} b & -1 \\ c & -1 \end{vmatrix} + 1 \begin{vmatrix} b & 3 \\ c & 4 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
- For the first determinant:
\[ \begin{vmatrix} 3 & -1 \\ 4 & -1 \end{vmatrix} = (3)(-1) - (4)(-1) = -3 + 4 = 1 \]
- For the second determinant:
\[ \begin{vmatrix} b & -1 \\ c & -1 \end{vmatrix} = (b)(-1) - (c)(-1) = -b + c = c - b \]
- For the third determinant:
\[ \begin{vmatrix} b & 3 \\ c & 4 \end{vmatrix} = (b)(4) - (c)(3) = 4b - 3c \]

3. **Substituting Back**:
Now substitute these back into the determinant equation:
\[ a(1) - 2(c - b) + (4b - 3c) = 0 \]
Simplifying this gives:
\[ a - 2c + 2b + 4b - 3c = 0 \]
Combining like terms:
\[ a + 6b - 5c = 0 \]

4. **Rearranging**:
Rearranging gives:
\[ a + 6b = 5c \]

5. **Conclusion**:
From the equation \(a + 6b = 5c\), we can express the relationship between \(a\), \(b\), and \(c\). This indicates that \(a\), \(b\), and \(c\) are in Arithmetic Progression (AP).