Prove that the equation $x^{2} + 6 x y + 9 y^{2} + 4 x + 12 y - 5 = 0$ represents two parallel lines.

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Solution

$x^{2} + 6 x y + 9 y^{2} + 4 x + 12 y - 5 = 0$

Comparing the equation with the general equation of second degree gives

$a = 1, b = 9, h = 3, g = 2, f = 6, c = - 5$

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

$tan θ = ∣ ∣ ∣ \frac{2 \sqrt{9 - 1 \times 9}}{1 + 9} ∣ ∣ ∣ = \frac{0}{9} tan θ = 0 \Rightarrow θ = {tan}^{-} (0) = 0^{\circ}$

Angle between the pair of straight lines is zero therefore the lines are parallel.
Hence proved

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