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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,we consider the homogeneous part of Eq. $(i)$: $1

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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,we consider the homogeneous part of Eq. $(i)$: $12x^2+25xy+12y^2 = 0$ $\Rightarrow (3x+4y)(4x+3y) = 0$ Let the lines be represented by $(3x+4y+c_1)(4x+3y+c_2) = 0$. Expanding this,we get: $12x^2+25xy+12y^2+(4c_1+3c_2)x+(3c_1+4c_2)y+c_1c_2 = 0$ Comparing this with Eq. $(i)$: $4c_1+3c_2 = 10$ ... (ii) $3c_1+4c_2 = 11$ ... (iii) $c_1c_2 = 2$ ... (iv) Solving (ii) and (iii): Adding (ii) and (iii): $7(c_1+c_2) = 21 \Rightarrow c_1+c_2 = 3$ Subtracting (iii) from (ii): $c_1-c_2 = -1$ Adding these: $2c_1 = 2 \Rightarrow c_1 = 1$. Then $c_2 = 2$. Check: $c_1c_2 = 1 	imes 2 = 2$,which satisfies (iv). The lines are $3x+4y+1=0$ and $4x+3y+2=0$. The perpendicular distances from the origin $(0,0)$ to these lines 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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