Find the value of k so that the following equations may represent pairs of straight lines :
$6 x^{2} + x y + k y^{2} - 11 x + 43 y - 35 = 0.$

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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 $6 x^{2} + x y + k y^{2} - 11 x + 43 y - 35 = 0$

$a = 6, b = k, h = \frac{1}{2}, g = \frac{- 11}{2}, f = \frac{43}{2}, c = - 35 Δ = (6 \times k \times - 35) + (2 \times \frac{43}{2} \times \frac{- 11}{2} \times \frac{1}{2}) - (6 \times \frac{43}{2} \times \frac{43}{2}) - (k \times \frac{- 11}{2} \times \frac{- 11}{2}) - (- 35 \times \frac{1}{2} \times \frac{1}{2}) = 0 - 210 k - \frac{473}{4} - \frac{11094}{4} - \frac{121 k}{4} + \frac{35}{4} = 0 - 840 k - 121 k - 473 - 11094 + 35 = 0 - 961 k - 11532 = 0 k = \frac{- 11532}{961} = - 12$

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