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Question

A line through A(5,4) meets the line x+3y+2=0,2x+y+4=0 and xy5=0 at the points B,C & D respectively, if (15AB)2+(10AC)2=(6AD)2
Let the equation of the line be lx+my+n=0.

Then find nml8.

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Solution

AB=r1, AC=r2, AD=r3

Using parametric form of line, we get

B(5+r1cosθ,4+r1sinθ)

It lie on line x+3y+2=0

AB=r1=512+2cosθ+3sinθ=15cosθ+3sinθ

Similarly,

r2=102cosθ+sinθ & r3=6cosθsinθ

Now,

(15AB)2+(10AC)2=(6AD)2

(cosθ+3sinθ)2+(2cosθ+sinθ)2=(cosθsinθ)2

4cos2θ+9sin2θ+12sinθcosθ=0

(2cosθ+3sinθ)3=0

tanθ=23

Equation of line through A is (y+4)=23(x+5)

2x+3y+22=0

l=2,m=3,n=22

nml=2223=17


240940_258542_ans_cf80ba33baef46668be3d94e34d26747.png

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