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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.
Solution
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.
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