Prove that the following equation represent two straight lines; find also their point of intersection and the angle between them.
$y^{2} + x y - 2 x^{2} - 5 x - y - 2 = 0$

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Solution

The general eqaution 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 $y^{2} + x y - 2 x^{2} - 5 x - y - 2 = 0$

$a = - 2, b = 1, h = \frac{1}{2}, g = \frac{- 5}{2}, f = \frac{- 1}{2}, c = - 2 Δ = (- 2 \times 1 \times - 2) + (2 \times \frac{- 1}{2} \times \frac{- 5}{2} \times \frac{1}{2}) - (- 2 \times \frac{- 1}{2} \times \frac{- 1}{2}) - (1 \times \frac{- 5}{2} \times \frac{- 5}{2}) - (- 2 \times \frac{1}{2} \times \frac{1}{2}) Δ = 4 + \frac{5}{4} + \frac{1}{2} - \frac{25}{4} + \frac{1}{2} = 0 Δ = 0$

$∴$ the equation represents a pair of straight lines. The point of intersection is found by partially differentiating the equation first with respect to $x$ and then with respect to $y$ and solving both the equations

$\frac{\partial}{\partial x} (y^{2} + x y - 2 x^{2} - 5 x - y - 2 = 0) 0 + y - 4 x - 5 - 0 - 0 = 0 4 x - y + 5 = 0.... (i) \frac{\partial}{\partial y} (y^{2} + x y - 2 x^{2} - 5 x - y - 2 = 0) 2 y + x - 0 - 0 - 1 - 0 = 0$
$x + 2 y - 1 = 0$ .... $(i i)$

Solving $(i)$ and $(i i)$

we get $x = - 1, y = 1$

So the point of intersection is $(- 1, 1)$

Angle between a pair of straight lines that is $tan θ = ∣ ∣ ∣ ∣ \frac{2 \sqrt{h^{2} - a b}}{a + b} ∣ ∣ ∣ ∣$

$tan θ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \frac{2 \sqrt{{(\frac{1}{2})}^{2} - (- 2) (1)}}{- 2 + 1} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ = 2 \times \frac{3}{2} = 3 \Rightarrow θ = {tan}^{- 1} (3)$

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