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Suppose a, b, c are sides of a scalene triangle. Let Δ=a    b    cb    c    ac    a    b .Then

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a
Δ≤0
b
Δ<0
c
Δ>0
d
Δ≥0

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detailed solution

Correct option is B

Using C1→C1+C2+C3 we getΔ=(a+b+c)Δ1where  Δ1=1    b    c1    c    a1    a    bApplying R2→R2−R1,R3→R3−R1 we getΔ1=1bc0c−ba−c0a−bb−c=−(b−c)2−(a−b)(a−c)=−a2+b2+c2−bc−ca−ab⇒        Δ1=−12(b−c)2+(c−a)2+(a−b)2<0 As a+b+c>0, we getΔ=(a+b+c)Δ1<0


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