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Question

A line passing through A(5,4) meets the line x+3y+2=0,2x+y+4=0 and x5y=0 at B,C and D respectively. If (15AB)2+(10AC)2=(6AD)2, then the equation of line is

A
2x+3y+22=0
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B
5x+4y+7=0
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C
3x2y+3=0
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D
None of these
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Solution

The correct option is A 2x+3y+22=0
Equation of line passing through A(5,4) and making an angle θ with x axis is
x+5cosθ=y+4sinθ=rx=5+rcosθ,y=4+rsinθ
So, parametric coordinates of B is x=5+ABcosθ,y=4+ABsinθ
Line AB,
x+3y+2=05+ABcosθ+3(4+ABsinθ)+2=015AB=cosθ+3sinθ(1)

similarly for line AC,
2x+y+4=02(5+ACcosθ)+(4+ACsinθ)+4=010AC=2cosθ+sinθ(2)

for line AD,
x5y=0(5+ADcosθ)5(4+ADsinθ)=06AD=cosθsinθ(3)

(15AB)2+(10AC)2=(6AD)2
Using equation (1),(2) and (3),
cos2θ+9sin2θ+6sinθcosθ+4cos2θ+sin2θ+4sinθcosθ =cos2θ+sin2θ2sinθcosθ(2cosθ+3sinθ)2=02cosθ+3sinθ=0tanθ=23

Therefore the equation of line is
y+4=23(x+5)3y+12=2x102x+3y+22=0

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