Answer :

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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