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Q.

Let 2a+3b+c=0. If  x+5y+7=0 and L = 0 are angle bisectors of two lines L1=0,L2=0 and L=0 is a member in the family of lines  ax+by+c=0 then area of triangle formed by L = 0 with coordinate axes is (in sq.units)

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a

49

b

492

c

495

d

4910

answer is D.

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Detailed Solution

Point of concurrence of ax+by+c=0 is (2, 3)Slope of L is −1−15=5Equation of line L = 0 is  y-3 = 5 (x-2)⇒5x-y-7=0Area of the triangle =(7)22|5(−1)|=4910sq.units
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