Let $p_{1}, p_{2}, p_{3}$ be the perpendicular distance between pair of parallel lines represented by $x^{2} - 3 x - 4 = 0,$ $y^{2} - 5 y + 6 = 0, 4 x^{2} + 20 x y + 25 y^{2} = 0$

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$p_{3} < p_{2} < p_{1}$

$p_{3} < p_{1} < p_{2}$

$p_{2} < p_{1} < p_{3}$

$p_{1} < p_{2} < p_{3}$

answer is A.

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Detailed Solution

$x^{2} - 3 x - 4 = 0 \Rightarrow x_{1} + x_{2} = \frac{- b}{a} = 3$ and

$x_{1} x_{2} = \frac{c}{a} = - 4 \Rightarrow p_{1} = |x_{1} - x_{2}| = \sqrt{{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2}} = 5$

$y^{2} - 5 y + 6 = 0 \Rightarrow y_{1} + y_{2} = 5, y_{1} y_{2} = 6$

$\Rightarrow p_{2} = |y_{1} - y_{2}| = \sqrt{25 - 24} = 1$

and $4 x^{2} + 20 x y + 25 y^{2} = 0$

$\Rightarrow (2 x + 5 y)^{2} = 0$

Lines are coincident $⊥^{lar}$ distance $(p_{3}) = 0$

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