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(B) Given the equation of the pair of straight lines: $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ $(i)$ First,factorize the homogeneous part: $12x^2 +

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$\frac{2}{25}$

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(B) Given the equation of the pair of straight lines: $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ $(i)$ First,factorize the homogeneous part: $12x^2 + 25xy + 12y^2 = (3x + 4y)(4x + 3y) = 0$. Let the two lines be $(3x + 4y + c_1) = 0$ and $(4x + 3y + c_2) = 0$. Their product is $(3x + 4y + c_1)(4x + 3y + c_2) = 12x^2 + 25xy + 12y^2 + (4c_1 + 3c_2)x + (3c_1 + 4c_2)y + c_1c_2 = 0$. Comparing this with equation $(i)$: $4c_1 + 3c_2 = 10$ $3c_1 + 4c_2 = 11$ Solving these equations: Multiply the first by $4$ and second by $3$: $16c_1 + 12c_2 = 40$ $9c_1 + 12c_2 = 33$ Subtracting gives $7c_1 = 7 \Rightarrow c_1 = 1$. Substituting $c_1 = 1$ into $4(1) + 3c_2 = 10$ $\Rightarrow 3c_2 = 6$ $\Rightarrow c_2 = 2$. The lines are $3x + 4y + 1 = 0$ and $4x + 3y + 2 = 0$. The perpendicular distances from the origin $(0,0)$ are: $p_1 = \frac{|0 + 0 + 1|}{\sqrt{3^2 + 4^2}} = \frac{1}{5}$ $p_2 = \frac{|0 + 0 + 2|}{\sqrt{4^2 + 3^2}} = \frac{2}{5}$ The product of the distances is $p_1 \cdot p_2 = \frac{1}{5} \cdot \frac{2}{5} = \frac{2}{25}$.

The product of the perpendicular distances from the origin to the pair of straight lines $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ is

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