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(A) The family of lines passing through the intersection of $L_1: x - 3y + 1 = 0$ and $L_2: 2x + 5y - 9 = 0$ is given by $(x - 3y + 1) + \lambda(2x +

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$2x + y - 5 = 0$

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(A) The family of lines passing through the intersection of $L_1: x - 3y + 1 = 0$ and $L_2: 2x + 5y - 9 = 0$ is given by $(x - 3y + 1) + \lambda(2x + 5y - 9) = 0$.This simplifies to $(1 + 2\lambda)x + (5\lambda - 3)y + (1 - 9\lambda) = 0$.The perpendicular distance from the origin $(0, 0)$ to this line is given as $\sqrt{5}$.Using the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have $\frac{|1 - 9\lambda|}{\sqrt{(1 + 2\lambda)^2 + (5\lambda - 3)^2}} = \sqrt{5}$.Squaring both sides: $\frac{(1 - 9\lambda)^2}{(1 + 2\lambda)^2 + (5\lambda - 3)^2} = 5$.$(1 - 81\lambda + 81\lambda^2) = 5(1 + 4\lambda + 4\lambda^2 + 25\lambda^2 - 30\lambda + 9)$.$(1 - 18\lambda + 81\lambda^2) = 5(29\lambda^2 - 26\lambda + 10) = 145\lambda^2 - 130\lambda + 50$.$64\lambda^2 - 112\lambda + 49 = 0$.$(8\lambda - 7)^2 = 0$,which gives $\lambda = \frac{7}{8}$.Substituting $\lambda = \frac{7}{8}$ into the equation: $(x - 3y + 1) + \frac{7}{8}(2x + 5y - 9) = 0$.$8x - 24y + 8 + 14x + 35y - 63 = 0$.$22x + 11y - 55 = 0$.Dividing by $11$,we get $2x + y - 5 = 0$.

Find the equation of the line passing through the intersection of the lines $x - 3y + 1 = 0$ and $2x + 5y - 9 = 0$ and whose distance from the origin is $\sqrt{5}$.

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