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Prabhat Aditya Grade: 12
        A line L drawn from pt. p(4,3) to meet the line L1:3x+4y+5=0,,,,and L2: 3x+4y+15=0.....at points A and B respectively .From point A a line perpendicular to L is drawn meeting the lines L2 at A1.Similarly from pt. P line perpendicular toL is drawn meeting the line L1 at B1.Thus a llgram AA1BB1 is completed.Find the equation of L so that area of the llgram is least?
8 years ago

Answers : (1)

Ramesh V
70 Points
										


AA1BB1 is a parallogram with 2 equal triangles AA1B and ABB1


Let AB=b


perpendicular distance btn L1 & L2 lines is p=2 units


Now consider traingle ABB1 ( from simple trigonometry )


we have its area = b2p / 2*(b2 - p2)1/2


for PA=r1 and PB=r2 and let  tan X be slope of line L


In polar coordinate form :


A ( 4+r1cos X , 3+r1sin X )   and B ( 4+r2cos X , 3+r2sin X )


on substituting A and B in L1 and L2 , we have


r1=|-29| / (3 cos X + 4 sin X)    and


r2=|-39| / (3 cos X + 4 sin X)


So AB= |r2 - r1|


AB = b =10 / (3 cos X + 4 sin X)


            b = 2 / sin (X+k)               :  where k=sin-1(3/5)


So total area is twice area of traingle ABB1


Area = b2p / (b2 - p2)1/2


          = 4 / sin (X+k)*cos (X+k)


          = 4 / sin 2(X+k)


For area to min. = sin 2(X+k) = 1


or                            2(X+k) = Pi /2


                               X = Pi/4 - k


     tan X = (1 - tan k)/(1+ tan k)


                = (1 -0.75)/(1+ 0.75) 


slope of L is : tan X = 1/7


So eqn. L is : x -7y+17 = 0




8 years ago
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