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If the equation of the locus of a point equidistant from the point (a_1, b_1 ) and (a_2, b_2 ) is (a_1-b_2 ) x+ (a_1-b_2 ) y+c=0, then the value of…

JEE Main 2003 — Mathematics Coordinate Geometry

2003mcqmedium

If the equation of the locus of a point equidistant from the point (a1,b1)\left(a_1, b_1\right)(a1​,b1​) and (a2,b2)\left(a_2, b_2\right)(a2​,b2​) is (a1−b2)x+(a1−b2)y+c=0\left(a_1-b_2\right) x+\left(a_1-b_2\right) y+c=0(a1​−b2​)x+(a1​−b2​)y+c=0, then the value of 'c' is

Official previous-year question

Held on 30 Apr 2003 · Verified 6 Jul 2026.

Options

  1. A

    a12+b12−a22−b22\sqrt{\mathrm{a}_1^2+\mathrm{b}_1^2-\mathrm{a}_2^2-\mathrm{b}_2^2}a12​+b12​−a22​−b22​​

  2. B

    12a22+b22−a12−b12\frac{1}{2} a_2^2+b_2^2-a_1^2-b_1^221​a22​+b22​−a12​−b12​

  3. C

    a12−a22+b12−b22\mathrm{a}_1^2-\mathrm{a}_2^2+\mathrm{b}_1^2-\mathrm{b}_2^2a12​−a22​+b12​−b22​

  4. D

    12(a12+a22+b12+b22)\frac{1}{2}\left(a_1^2+a_2^2+b_1^2+b_2^2\right)21​(a12​+a22​+b12​+b22​)

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