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A line L is drawn from P(4,3) to meet the lines L1: 3x+4y+5=0 and L2: 3x+4y+15=0 at point 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 AA1BB1 is formed. Find the equation of L so that the area of the parallelogram AA1BB1 is least.

A line L is drawn from P(4,3) to meet the lines L1: 3x+4y+5=0 and L2: 3x+4y+15=0 at point 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 AA1BB1 is formed. Find the equation of L so that the area of the parallelogram AA1BB1 is least.

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1 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago

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