# If $$\lambda x^2 – 10xy + 12y^2 + 5x – 16y – 3$$ = 0 represents a pair of straight lines, then $$\lambda$$ is equal to

## Solution :

Comparing with $$ax^2+2hxy+by^2+2gx+2fy+c$$ = 0

Here a = $$\lambda$$, b = 12, c = -3, f = -8, g = 5/2, h = -5

Using condition $$abc+2fgh-af^2-bg^2-ch^2$$ = 0, we have

$$\lambda$$(12)(-3) + 2(-8)(5/2)(-5) – $$\lambda$$(64) – 12(25/4) + 3(25) = 0

$$\implies$$  -36$$\lambda$$ + 200 – 64$$\lambda$$ – 75 + 75 = 0

$$\implies$$  100$$\lambda$$ = 200

$$\therefore$$  $$\lambda$$ = 2

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