Question

If a, b, c are in A.P., prove that the straight lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent.

Answer 1

The given lines can be written as follows:
ax + 2y + 1 = 0           ... (1)
bx + 3y + 1 = 0           ... (2)
cx + 4y + 1 = 0           ... (3)
Consider the following determinant.

\[\begin{vmatrix}a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1\end{vmatrix}\]

Applying the transformation \[R_1 \to R_1 - R_2 \text { and } R_2 \to R_2 - R_3\],

\[\begin{vmatrix}a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1\end{vmatrix} = \begin{vmatrix}a - b & - 1 & 0 \\ b - c & - 1 & 0 \\ c & 4 & 1\end{vmatrix}\] 

\[\Rightarrow \begin{vmatrix}a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1\end{vmatrix} = \left( - a + b + b - c \right) = 2b - a - c\]

Given:
2b = a + c

\[\begin{vmatrix}a & 2 & 1 \\ b & 3 & 1 \\ c & 4 & 1\end{vmatrix} = a + c - a - c = 0\]

Hence, the given lines are concurrent, provided 2b = a + c.