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 C1→C1 + C2,
⇒
⇒ ⇒
Hence proved, the given lines are concurrent.