If the equation of the locus of a point equidistant from the points $({{a}_{1}},{{b}_{1}})$ and $({{a}_{2}},{{b}_{2}})$ is $({{a}_{1}}-{{a}_{2}})x+({{b}_{1}}-{{b}_{2}})y+c=0$, then the value of c is

Question

Accepted Answer

$\frac{1}{2}(a_{2}^{2}+b_{2}^{2}-a_{1}^{2}-b_{1}^{2})$

Answer

$a_{1}^{2}-a_{2}^{2}+b_{1}^{2}-b_{2}^{2}$

Answer

$\sqrt{a_{1}^{2}+b_{1}^{2}-a_{2}^{2}-b_{2}^{2}}$

Answer

$\frac{1}{2}(a_{1}^{2}+a_{2}^{2}+b_{1}^{2}+b_{2}^{2})$

coordinate

Question: If the equation of the locus of a point equidistant from the points $({{a}_{1}},{{b}_{1}})$ and $({{a}_{2}},{{b}_{2}})$ is $({{a}_{1}}-{{a}_{2}})x+({{b}_{1}}-{{b}_{2}})y+c=0$, then the value of c is

coordinate

Question: If the equation of the locus of a point equidistant from the points $({{a}_{1}},{{b}_{1}})$ and $({{a}_{2}},{{b}_{2}})$ is $({{a}_{1}}-{{a}_{2}})x+({{b}_{1}}-{{b}_{2}})y+c=0$, then the value of c is

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