Find the value of k so that the following equations may represent pairs of straight lines :
$x^{2} + k x y + y^{2} - 5 x - 7 y + 6 = 0$

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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 $Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$

$x^{2} + k x y + y^{2} - 5 x - 7 y + 6 = 0$

$a = 1, b = 1, h = \frac{k}{2}, g = \frac{- 5}{2}, f = \frac{- 7}{2}, c = 6$
$Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$Δ = (1 \times 1 \times 6) + (2 \times \frac{- 7}{2} \times \frac{- 5}{2} \times \frac{k}{2}) - (1 \times \frac{- 7}{2} \times \frac{- 7}{2}) - (1 \times \frac{- 5}{2} \times \frac{- 5}{2}) - (6 \times \frac{k}{2} \times \frac{k}{2}) = 0 Δ = 6 + \frac{35 k}{4} - \frac{49}{4} - \frac{25}{4} - \frac{3 k^{2}}{2} = 0$

$24 + 35 k - 49 - 25 - 6 k^{2} = 0 6 k^{2} - 35 k + 50 = 0 6 k^{2} - 15 k - 20 k + 50 = 0 3 k (2 k - 5) - 10 (2 k - 5) = 0 (3 k - 10) (2 k - 5) = 0 \Rightarrow k = \frac{10}{3}, \frac{5}{2}$

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