Guest

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?

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?

Grade:12

1 Answers

Ramesh V
70 Points
12 years ago

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


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free