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Question

If ∣ ∣ ∣ab+ca2bc+ab2ca+bc2∣ ∣ ∣=0, where a, b, c are distinct real numbers, then the straight line ax + by + c = 0 passes through the fixed point



A

(1, -1)

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B

(-1, -1)

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C

(0, 1)

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D

(1, 1)

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Solution

The correct option is D

(1, 1)


Δ=(a+b+c)×∣ ∣ ∣1b+ca21c+ab21a+bc2∣ ∣ ∣
(Applying C1C1+C2 and taking (a + b + c) common)
=(a+b+c)∣ ∣ ∣1b+ca20ab(ba)(b+a)0ac(ca)(c+a)∣ ∣ ∣(R2R2R1,R3R3R1)
=(a+b+c)(ab)(ca)∣ ∣ ∣1b+ca201(b+a)01(ca)∣ ∣ ∣
= (a + b + c) (a - b) (c - a) [c + a - b -a]
= -(a + b + c) (a - b) (b - c) (c - a)

Since a, b, c are distinct real numbers then Δ=0 only when a + b + c = 0.
Hence line ax + by + c = 0 passes through the fixed point (1, 1)


Mathematics

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