Find the value of k so that the following equations may represent pairs of straight lines :
$6 x^{2} + 11 x y - 10 y^{2} + x + 31 y + k = 0.$

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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 ${6 x}^{2} + 11 x y - 10 y^{2} + x + 31 y + k = 0$

$a = 6, b = - 10, h = \frac{11}{2}, g = \frac{1}{2}, f = \frac{31}{2}, c = k Δ = (6 \times - 10 \times k) + (2 \times \frac{31}{2} \times \frac{1}{2} \times \frac{11}{2}) - (6 \times \frac{31}{2} \times \frac{31}{2}) - (- 10 \times \frac{1}{2} \times \frac{1}{2}) - (k \times \frac{11}{2} \times \frac{11}{2}) = 0 - 60 k + \frac{341}{4} - \frac{5776}{4} + \frac{10}{4} - \frac{121 k}{4} = 0 \frac{- 361 k}{4} - \frac{5415}{4} = 0 \Rightarrow k = - 15$

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