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Q.) A line 'L' is drawn from P(4, 3) to meet the lines L1 and L2 given by 3x + 4y + 5 = 0 and 3x + 4y + 15 = 0 at points A and B, respectively. From 'A', a line perpendicular to L is drawn meeting the line L2 at A1'. Similarly from point 'B', a line perpendicular to L is drawn meeting the line L1 at B1'. Thus a parallelogram AA1'BB1' is formed. Then the equation of 'L' so that the area of the parallelogram AA1'BB1' is least is
[A]x - 7y + 17 = 0
[B]7x + y + 31 = 0
[C]x - 7y - 17 = 0
[D]x + 7y - 31 = 0