MathematicsMedium128×since 2002Q5563If 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−a2)x+(b1−b2)y+c=0\left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + c = 0(a1−a2)x+(b1−b2)y+c=0 , then the value of ′c′'c'′c′ is :Aa12+b12−a22−b22\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2}a12+b12−a22−b22B12(a22+b22−a12−b12){1 \over 2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)21(a22+b22−a12−b12)Ca12−a22+b12−b22{{a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2}a12−a22+b12−b22D12(a12+a22+b12+b22){1 \over 2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)21(a12+a22+b12+b22).Check answerSkip