If the equation of the locus point equidistant from the points $\left({\mathrm{a}}_1,{\mathrm{b}}_1\right)$ and $\left({\mathrm{a}}_2,{\mathrm{b}}_2\right)$ is $\left({\mathrm{a}}_1-{\mathrm{a}}_2\right)\mathrm{x}+\left({\mathrm{b}}_1-{\mathrm{b}}_2\right)\mathrm{y}+\mathrm{c}=0,$ then the value of c is 

Let $\left(\mathrm{h},\mathrm{k}\right)$ be the points on the locus. Then by the given conditions,

${\left(\mathrm{h}-\mathrm{a}\right)}^2+{\left(\mathrm{k}-{\mathrm{b}}_1\right)}^2={\left(\mathrm{h}-{\mathrm{a}}_2\right)}^2+{\left(\mathrm{k}-{\mathrm{b}}_2\right)}^2$

or $2\mathrm{h}\left({\mathrm{a}}_1-{\mathrm{a}}_2\right)+2\mathrm{k}\left({\mathrm{b}}_1-{\mathrm{b}}_2\right)+{\mathrm{a}}_2^2+{\mathrm{b}}_2^2-{\mathrm{a}}_1^2-{\mathrm{b}}_1^2=0$

or $2\mathrm{h}\left({\mathrm{a}}_1-{\mathrm{a}}_2\right)+\mathrm{k}\left({\mathrm{b}}_1-{\mathrm{b}}_2\right)+\frac{1}{2}\left({\mathrm{a}}_2^2+{\mathrm{b}}_2^2-{\mathrm{a}}_1^2-{\mathrm{b}}_1^2\right)=0$         …(1)

Also since $\left(\mathrm{h},\mathrm{k}\right)$ lies on the given locus, we have 

$\left({\mathrm{a}}_1-{\mathrm{a}}_2\right)\mathrm{h}+\mathrm{b}\left({\mathrm{b}}_1-{\mathrm{b}}_2\right)\mathrm{k}+\mathrm{c}=0$                                         …(2)

Comparing (1) and (2) , we get 

$\mathrm{c}=\frac{1}{2}\left({\mathrm{a}}_2^2+{\mathrm{b}}_2^2-{\mathrm{a}}_1^2-{\mathrm{b}}_1^2\right)$