If the equation of the locus of a point equidistant from the point (a 1, b₁) and (a 2, b₂) is (a₁ - a₂)x + (b₁ - b₂)y + c = 0, then the value of 'c' is:

(B) 1/2(a₂² + b₂² - a₁² - b₁²)

(A) √(a₁² + b₁² - a₂² - b₂²)

(C) a₁² - a₂² + b₁² - b₂²

(D) 1/2(a₁² + a₂² + b₁² + b₂²).

Coordinate Geometry: JEE Main 2003 Mathematics question

JEE Main 2003 · Annual, Paper 1 · Q5563

JEE Main 2003Annual, Paper 1MathematicsMediumCoordinate Geometry478× since 2002

If the equation of the locus of a point equidistant from the point

\left( {{a_{1,}}{b_1}} \right)

and

\left( {{a_{2,}}{b_2}} \right)

\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0

, then the value of

'c'

is :

JEE Main 2003 · Annual, Paper 1 · Q5563

JEE Main 2003Annual, Paper 1MathematicsMediumCoordinate Geometry478× since 2002

If the equation of the locus of a point equidistant from the point

\left( {{a_{1,}}{b_1}} \right)

and

\left( {{a_{2,}}{b_2}} \right)

\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0

, then the value of

'c'

is :