# 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.

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 C1C1 + C2,

Hence proved, the given lines are concurrent.

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