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(B) The equation of a line passing through the intersection of $2x + y - 4 = 0$ and $x - 3y + 5 = 0$ is given by the family of lines equation: $(2x +

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$x + 2y - 5 = 0$

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(B) The equation of a line passing through the intersection of $2x + y - 4 = 0$ and $x - 3y + 5 = 0$ is given by the family of lines equation: $(2x + y - 4) + \lambda(x - 3y + 5) = 0$ ... $(i)$ Rearranging the terms,we get: $x(2 + \lambda) + y(1 - 3\lambda) + (5\lambda - 4) = 0$. The perpendicular distance from the origin $(0, 0)$ to this line is $\sqrt{5}$. Using the distance formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$,we have: $\left|\frac{5\lambda - 4}{\sqrt{(2 + \lambda)^2 + (1 - 3\lambda)^2}}\right| = \sqrt{5}$ Squaring both sides: $\frac{(5\lambda - 4)^2}{4 + \lambda^2 + 4\lambda + 1 + 9\lambda^2 - 6\lambda} = 5$ $\frac{(5\lambda - 4)^2}{10\lambda^2 - 2\lambda + 5} = 5$ $25\lambda^2 - 40\lambda + 16 = 50\lambda^2 - 10\lambda + 25$ $25\lambda^2 + 30\lambda + 9 = 0$ $(5\lambda + 3)^2 = 0 \Rightarrow \lambda = -\frac{3}{5}$. Substituting $\lambda = -\frac{3}{5}$ into equation $(i)$: $(2x + y - 4) - \frac{3}{5}(x - 3y + 5) = 0$ $5(2x + y - 4) - 3(x - 3y + 5) = 0$ $10x + 5y - 20 - 3x + 9y - 15 = 0$ $7x + 14y - 35 = 0$ Dividing by $7$,we get: $x + 2y - 5 = 0$.

The equation of the line passing through the point of intersection of the lines $2x + y - 4 = 0$ and $x - 3y + 5 = 0$ and lying at a distance of $\sqrt{5}$ units from the origin is:

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