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Q.

If x - 2y + 4 - 0 and 2x + y - 5 = 0 are the sides of an isosceles triangle having area 10 sq. units, then possible equation of the third side is

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a

x+3y=-1

b

x+3y=19

c

3x-y=-9

d

3x-y=11

answer is A, B, C, D.

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Detailed Solution

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Given lines are mutually perpendicular and intersect at  A(65,135)

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Equation of angle bisectors. B1, and B2 of the given lines are
 x2y+4=±(2x+y5)or  x+3y=9 and 3xy=1
Side BC will be parallel to these bisectors.

Let AD = a.

 AB=a2

Area of triangle ABC=12×(a2)2=a2  
 a2=10or  a=10
Let equation of BC bex+3y=λ. Then

 10=|65+395λ|10 λ=1,19

So, equation of BC is x+3y=1 or x+3y=19
If equation of BC is x+3y=λ, then

 10=|185135λ|10 λ=9,11

Therefore, equation of BC is 3xy=9 or 3xy=11

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If x - 2y + 4 - 0 and 2x + y - 5 = 0 are the sides of an isosceles triangle having area 10 sq. units, then possible equation of the third side is