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JEE Main2003MathematicsStraight LinesActual

If the equation of the locus of a point equidistant from the point (a₁, b₁ ) and (a₂, b₂ ) is (a₁-b₂ ) x+ (a₁-b₂ ) y+c=0 , then the value of 'c' is

Options

  1. Aa ₁^2+ b ₁^2- a ₂^2- b ₂^2
  2. B1 2 a₂^2+b₂^2-a₁^2-b₁^2
  3. Ca ₁^2- a ₂^2+ b ₁^2- b ₂^2
  4. D1 2 (a₁^2+a₂^2+b₁^2+b₂^2 )

Correct answer

B. 1 2 a₂^2+b₂^2-a₁^2-b₁^2

Step-by-step solution

(h-a₁ )^2+ (k-b₁ )^2= (h-a₂ )^2+ (k-b₂ )^2 (a₁-a₂ ) x+ (b₁-b₂ ) y+ 1 2 (a₂^2+b₂^2-a₁^2-b₁^2 )=0C= 1 2 (a₂^2+b₂^2-a₁^2-b₁^2 )

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