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Suppose a, b and c are distinct real numbers. Let Δ=aa+cabbcba+bc+bcac=0 .Then the straight line a(x5)+b(y2)+c=0 passes through the fixed point

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a
(5, 2)
b
(6, 2)
c
(6, 3)
d
(5, 3)

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detailed solution

Correct option is C

Applying C1→aC1+bC2+cC3 we getLet          Δ=1aa(a+b+c)a+ca−bb(a+b+c)ba+bc(a+b+c)c−ac=1aa(a+b+c)Δ1, whereΔ1=aa+ca−bbba+bcc−acUsing C2→C2−C1,C3→C3−C1 we getΔ1=ac−bb0ac−a0=aa2+b2+c2∴ Δ=(a+b+c)a2+b2+c2=0As a, b, c are distinct real numbers, a2+b2+c2≠0 thereforea+b+c=0⇒the line a(x–5)+b(y–2)+c=0 passes through (6, 3)


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