Show that the straight lines given by x(a + 2b) + y(a + 3b) = a + b fo

1.	Show that the straight lines given by x(a + 2b) + y(a + 3b) = a + b for different value of a and b Pass through a fixed point.
Answer» Given equation can be written as a(x + y – 1) + b(2x + 3y – 1) = 0 ⇒ (x + y − 1) + λ(2x + 3y − 1) = 0, where λ = \(\frac{b}{a}\). This is the form L₁ + λL₂ = 0. So, it represents that line passing through the intersection of x + y – 1 = 0 and 2x + 3y – 1 = 0 ∴ The straight lines pass through a fixed point.

Show that the straight lines given by x(a + 2b) + y(a + 3b) = a + b for different value of a and b Pass through a fixed point.

Answer»

Given equation can be written as

a(x + y – 1) + b(2x + 3y – 1) = 0

⇒ (x + y − 1) + λ(2x + 3y − 1) = 0, where λ = \(\frac{b}{a}\).

This is the form L₁ + λL₂ = 0. So, it represents that line passing through the intersection of

x + y – 1 = 0 and 2x + 3y – 1 = 0

∴ The straight lines pass through a fixed point.