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If 3a + 2b + 6c = 0, the family of straight lines ax + by + c = 0 passes through a fixed point whose coordinates are given by
  • a)
    (1/2, 1/3)
  • b)
    (2, 3)
  • c)
    (3, 2)
  • d)
    (1/3, 1/2)
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
If3a + 2b + 6c = 0, the family of straight linesax + by + c = 0 passes...
∵ 3a + 2b + 6c = 0
∴ ax + by + c = 0

⇒ 6ax + 6by − 3a − 2b = 0
⇒ 3a(2x − 1) + 2b(3y − 1) = 0

P + λQ = 0, ∴ P = 0, Q = 0
Hence, fixed points is (1/2,1/3)
Alternative Solution
If we put the point in the equation of line, it should lead to the condition
3a + 2b + 6c = 0. This happens only for (1/2, 1/3).
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If3a + 2b + 6c = 0, the family of straight linesax + by + c = 0 passes through a fixed point whose coordinates are given bya)(1/2, 1/3)b)(2, 3)c)(3, 2)d)(1/3, 1/2)Correct answer is option 'A'. Can you explain this answer?
Question Description
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