The distance between the pair of parallel lines represented by $x^{2} + 4 x y + 4 y^{2} + 3 x + 6 y - 4 = 0$ is ___ units

\sqrt{5}

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\frac{1}{\sqrt{5}}

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3

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\frac{1}{\sqrt{3}}

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Solution

The correct option is A $\sqrt{5}$
$x^{2} + 4 x y + 4 y^{2} + 3 x + 6 y - 4 = 0$
$\Rightarrow (x + 2 y)^{2} + 3 (x + 2 y) - 4 = 0$
$\Rightarrow ((x + 2 y) + 4) ((x + 2 y) - 1) = 0$
Therefore, the lines are
$x + 2 y + 4 = 0$
and $x + 2 y - 1 = 0$
Required distance $= \frac{| 4 - (- 1) |}{\sqrt{1 + 4}}$
$= \frac{5}{\sqrt{5}} = \sqrt{5}$ units

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