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(A) Let the family of lines passing through the intersection of $L_1: 2x - y + 2 = 0$ and $L_2: x + y + 4 = 0$ be given by $L_1 + \lambda L_2 = 0$.$(2

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$y + 2 = 0$

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(A) Let the family of lines passing through the intersection of $L_1: 2x - y + 2 = 0$ and $L_2: x + y + 4 = 0$ be given by $L_1 + \lambda L_2 = 0$.$(2x - y + 2) + \lambda(x + y + 4) = 0$.Since the line passes through the point $(5, -2)$,we substitute $x = 5$ and $y = -2$ into the equation:$(2(5) - (-2) + 2) + \lambda(5 + (-2) + 4) = 0$.$(10 + 2 + 2) + \lambda(5 - 2 + 4) = 0$.$14 + \lambda(7) = 0$.$7\lambda = -14$.$\lambda = -2$.Substituting $\lambda = -2$ back into the family equation:$(2x - y + 2) - 2(x + y + 4) = 0$.$2x - y + 2 - 2x - 2y - 8 = 0$.$-3y - 6 = 0$.$y + 2 = 0$.

The equation of the line passing through the point of intersection of lines $2x - y + 2 = 0$ and $x + y + 4 = 0$ and the point $(5, -2)$ is

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