Supposse that the distance from the point [tex](x,y)[/tex] to the point [tex](3,0)[/tex] is equal to the distance from [tex](x,y)[/tex] to the point [tex](7,0)[/tex]. Then, by the formula of the distnace we must have
[tex]\sqrt{(x-3)^2 + (y-0)^2}=\sqrt{(x-7)^2 + (y-0)^2}[/tex]
cancel the square root and the [tex](y-0)^2[/tex]'s, and then expand the parenthesis to obtain
[tex]x^2 - 6x + 9 = x^2 - 14x + 49[/tex]
then, simplifying we obtain
[tex]8x = 40 [/tex]
therfore we must have
[tex]x=5[/tex]
this means that the points satisfying the propertie must have first component equal to 5. So we can give a lot of examples of such points: [tex](5,0), (5,7),(5,1/2), (5,-10),...[/tex]. The set of this points give us a straight line and the points (3,0) and (7,0) are symmetric with respect to this line.