L is variable lin

Let ax + by + c = 0 be the variable line. It is given that the algebraic sum of the distances
of the points (1, 1), (2, 0) and (0, 2) from the line is equal to zero.

∴

⇒ 3a + 3b + 3c = 0

⇒ a + b + c = 0

Substituting c = – a – b in ax + by + c = 0, we get:

ax + by – a – b = 0

⇒ a(x – 1) + b(y – 1) = 0

⇒

This line is of the form L₁ + λL₂ = 0, which passes through the intersection of L₁ = 0 and L₂ = 0, i.e. x – 1 = 0 and y – 1 = 0.

⇒ x = 1, y = 1