Product of perpendicular distances drawn from origin to pair of straight lines $12 x^{2} + 25 x y + 12 y^{2} + 10 x + 11 y + 2 = 0$ is __ sq. units

\frac{1}{25}

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\frac{2}{25}

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\frac{3}{25}

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\frac{4}{25}

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Solution

The correct option is B $\frac{2}{25}$
$12 x^{2} + 25 x y + 12 y^{2} = 12 x^{2} + 16 x y + 9 x y + 12 y^{2} = (4 x + 3 y) (3 x + 4 y)$

So, the required pair of lines are
$4 x + 3 y + λ_{1} = 0 and 3 x + 4 y + λ_{2} = 0$
Therefore,
$(4 x + 3 y + λ_{1}) (3 x + 4 y + λ_{2}) = 12 x^{2} + 25 x y + 12 y^{2} + 10 x + 11 y + 2 \Rightarrow λ_{1} \cdot λ_{2} = 2$

Perpendicular distance from $(0, 0)$ to $4 x + 3 y + λ_{1} = ∣ ∣ ∣ \frac{λ_{1}}{5} ∣ ∣ ∣$ units
Perpendicular distance from $(0, 0)$ to $3 x + 4 y + λ_{2} = ∣ ∣ ∣ \frac{λ_{2}}{5} ∣ ∣ ∣$ units

So, product of perpendicular distances $= ∣ ∣ ∣ \frac{λ_{1} . λ_{2}}{25} ∣ ∣ ∣ = \frac{2}{25}$ sq units

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