A line passing through A(−5,−4) meets the line x+3y+2=0,2x+y+4=0 and x−5−y=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
3x−2y+3=0
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D
None of these
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Solution
The correct option is A2x+3y+22=0 Equation of line passing through A(−5,−4) and making an angle θ with x− axis is x+5cosθ=y+4sinθ=r⇒x=−5+rcosθ,y=−4+rsinθ So, parametric coordinates of B is x=−5+ABcosθ,y=−4+ABsinθ Line AB, x+3y+2=0⇒−5+ABcosθ+3(−4+ABsinθ)+2=0⇒15AB=cosθ+3sinθ⋯(1)
similarly for line AC, 2x+y+4=0⇒2(−5+ACcosθ)+(−4+ACsinθ)+4=0⇒10AC=2cosθ+sinθ⋯(2)
for line AD, x−5−y=0⇒(−5+ADcosθ)−5−(−4+ADsinθ)=0⇒6AD=cosθ−sinθ⋯(3)
(15AB)2+(10AC)2=(6AD)2 Using equation (1),(2) and (3), ⇒cos2θ+9sin2θ+6sinθcosθ+4cos2θ+sin2θ+4sinθcosθ=cos2θ+sin2θ−2sinθcosθ⇒(2cosθ+3sinθ)2=0⇒2cosθ+3sinθ=0⇒tanθ=−23
Therefore the equation of line is y+4=−23(x+5)⇒3y+12=−2x−10∴2x+3y+22=0