Find the L2 equation of the line L: x + 2Y + 1 = 0, which is symmetric with respect to the line L: 3x-y-1 = 0

It can be seen from the proposition that any point P (x, y) on the line L2, with respect to the symmetric point Q ((- 4x + 3Y + 3) / 5, (3x + 4y-1) / 5) of the line L: 3x-y-1 = 0, must be on the line L1: x + 2Y + 1 = 0, ■ [(- 4x + 3Y + 3) / 5] + [2 (3x + 4y-1) / 5] + 1 = 0

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