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Question

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

A
2x+3y+22=0
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B
5x4y+7=0
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C
3x2y+3=0
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D
2x+3y+7=0
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Solution

## The correct option is C 2x+3y+22=0The parametric form of equation of line passing through A(-5,-4) x+5cosθ=y+4sinθ=rLet AB=r1Then the coordinates of B are (r1cosθ−5,r1sinθ−4)Since, point B lies on the line x+3y+2=0⇒r1(3sinθ+cosθ)=15⇒15r1=3sinθ+cosθLet AC=r2Coordinates of C are (r2cosθ−5,r2sinθ−4)Since, point C lies on the line 2x+y+4=0⇒r2(sinθ+2cosθ)=10⇒10r2=sinθ+2cosθLet AD=r3Coordinates of D are (r3cosθ−5,r3sinθ−4)Since, point D lies on the line x−y−5=0⇒r3(−sinθ+cosθ)=6⇒6r3=−sinθ+cosθ(15AB)2+(10AC)2=(6AD)2(15r1)2+(10r2)2=(6r3)2⇒(3sinθ+cosθ)2+(sinθ+2cosθ)2=(−sinθ+cosθ)2 ⇒(3sinθ+2cosθ)2=0⇒tanθ=−23So, the equation of required line isy+4=−23(x+5)⇒2x+3y+22=0

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