Question

Show that the straight lines L₁ = (b + c) x + ay + 1 = 0, L₂ = (c + a) x + by + 1 = 0 and L₃ = (a + b) x + cy + 1 = 0 are concurrent.

Answer 1

The given lines can be written as follows:
(b + c) x + ay + 1 = 0           ... (1)
(c + a) x + by + 1 = 0           ... (2)
(a + b) x + cy + 1 = 0           ... (3)
Consider the following determinant.

\[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix}\]

Applying the transformation  \[C_1 \to C_1 + C_2\] \[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix} = \begin{vmatrix}a + b + c & a & 1 \\ c + a + b & b & 1 \\ a + b + c & c & 1\end{vmatrix}\]

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

\[\Rightarrow\] \[\begin{vmatrix}b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1\end{vmatrix}\] =0

Hence, the given lines are concurrent.