Find the value of k so that the following equations may represent pairs of straight lines :
$2 x^{2} + x y - y^{2} + k x + 6 y - 9 = 0$

Open in App

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 $2 x^{2} + x y - y^{2} + k x + 6 y - 9 = 0$

$a = 2, b = - 1, h = \frac{1}{2}, g = \frac{k}{2}, f = 3, c = - 9$
$Δ = a b c + 2 f g h - a f^{2} - b g^{2} - c h^{2} = 0$
$Δ = (2 \times - 1 \times - 9) + (2 \times 3 \times \frac{k}{2} \times \frac{1}{2}) - (2 \times 3 \times 3) - (- 1 \times \frac{k}{2} \times \frac{k}{2}) - (- 9 \times \frac{1}{2} \times \frac{1}{2}) = 0 18 + \frac{3 k}{2} - 18 + \frac{k^{2}}{4} + \frac{9}{4} \Rightarrow k^{2} + 6 k + 9 = 0 {(k + 3)}^{2} = 0 \Rightarrow k = - 3$

Suggest Corrections

Similar questions

- 2 x^{2} + x y + y^{2} - 5 x + y + K = 0

12 x^{2} - k x y + 2 y^{2} + 11 x - 5 y + 2 = 0.

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

Join BYJU'S Learning Program