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Question

# A line through A(-5, -4) meets the lines 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
23x + 14y - 40 = 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 A 23x + 14y - 40 = 0x+5cosθ=y+4sinθ=r1AB=r2AC=r3AD (r1cosθ−5,r1sinθ−4) lies on x+3y+2=0 ∴r1=15cosθ+3sinθ Similarly 10AC=2cosθ+sinθ and 6AD=cosθ−sinθ Putting in the given relation, we get (2cosθ+3sinθ)2=0 ∴tanθ=−23⇒y+4=−23(x+5)⇒2x+3y+22=0.

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