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Given:

L₁ = (b + c)x + ay + 1 = 0

L₂ = (c + a)x + by + 1 = 0

L₃ = (a + b)x + cy + 1 = 0

To prove:

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.

Concept Used:

If three lines are concurrent then determinant of equation is zero.

Explanation:

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.

Applying the transformation C₁→C₁ + C₂,

⇒

⇒ ⇒

Hence proved, the given lines are concurrent.