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The equation of the locus of the foot of perpendicular drawn from 5,6 on the family of lines x - 2 + λ y - 3 = 0 (where λ ∈ R ) is

Options

  1. Ax - 1 x - 3 + y - 2 y - 6 = 0
  2. Bx - 5 x - 6 + y - 2 y - 3 = 0
  3. Cx - 2 x - 5 + y - 3 y - 6 = 0
  4. Dx + 2 x + 5 + y + 3 y + 6 = 0

Correct answer

C. x - 2 x - 5 + y - 3 y - 6 = 0

Step-by-step solution

Let A = 5,6 and the point of concurrency of the family of lines x - 2 + λ y - 3 = 0 is 2,3 = B and foot of the perpendicular from A to the family of lines is P = h , k Now, P A is perpendicular to P B ⇒ s l o p e  o f  P A × s l o p e  o f  P B = - 1 ⇒ k - 6 h - 5 × k - 3 h - 2 = - 1 ⇒ k - 6 k - 3 = - h - 5 h - 2 ⇒ locus is x - 2 x - 5 + y - 3 y - 6 = 0

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