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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