Find the value of k so that the following equation may represent pairs of straight lines :
$12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k = 0.$

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Solution

The general equation of second degree $a x^{2} + 2 h x y + {b y}^{2} + 2 g x + 2 f y + c = 0$ Represents a pair of straight lines if $Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$

For $12 x^{2} - 10 x y + 2 y^{2} + 11 x - 5 y + k = 0$

$a = 12, b = 2, h = - 5, g = \frac{11}{2}, f = \frac{- 5}{2}, c = k Δ = (12 \times 2 \times k) + (2 \times \frac{- 5}{2} \times \frac{11}{2} \times - 5) - (12 \times \frac{- 5}{2} \times \frac{- 5}{2}) - (2 \times \frac{11}{2} \times \frac{11}{2}) - (k \times - 5 \times - 5) = 0 24 k + \frac{275}{2} - 75 - \frac{121}{2} - 25 k = 0 - k + 2 = 0 \Rightarrow k = 2$

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