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.

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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.

$|\begin{matrix} b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1 \end{matrix}|$

Applying the transformation $C_{1} \to C_{1} + C_{2}$ ,

$|\begin{matrix} b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1 \end{matrix}| = |\begin{matrix} a + b + c & a & 1 \\ c + a + b & b & 1 \\ a + b + c & c & 1 \end{matrix}|$

$\Rightarrow$ $|\begin{matrix} b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1 \end{matrix}|$ = $(a + b + c) |\begin{matrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{matrix}|$

$\Rightarrow$ $|\begin{matrix} b + c & a & 1 \\ c + a & b & 1 \\ a + b & c & 1 \end{matrix}|$ =0

Hence, the given lines are concurrent.

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