Question
If the equation of the locus of a point equidistant from the point $$\left( {{a_1},\,{b_1}} \right)$$ and $$\left( {{a_2},\,{b_2}} \right)$$ is $$\left( {{a_1} - \,{b_2}} \right)x + \left( {{a_1} - \,{b_2}} \right)y + c = 0,$$ then the value of $$'c\,'$$ is-
A.
$$\sqrt {{a_1}^2 + {b_1}^2 - {a_2}^2 - {b_2}^2} $$
B.
$$\frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$
C.
$${a_1}^2 - {a_2}^2 + {b_1}^2 - {b_2}^2$$
D.
$$\frac{1}{2}\left( {{a_1}^2 + {a_2}^2 + {b_1}^2 + {b_2}^2} \right)$$
Answer :
$$\frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right)$$
Solution :
$$\eqalign{
& {\left( {x - {a_1}} \right)^2} + {\left( {y - {b_1}} \right)^2} = {\left( {x - {a_2}} \right)^2} + {\left( {y - {b_2}} \right)^2} \cr
& \left( {{a_1} - {a_2}} \right)x + \left( {{b_1} - {b_2}} \right)y + \frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) = 0 \cr
& c = \frac{1}{2}\left( {{a_2}^2 + {b_2}^2 - {a_1}^2 - {b_1}^2} \right) \cr} $$