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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.

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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.

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