Find the equation of the locus of the point P equidistant from the points (2,3),(1,4)

Find the equation of the locus of the point P equidistant from the points (-3,2),(0,4)

Find the equation of the locus of the point "P" equidistant from the points (-3,2),(0,4)

The equation of the locus of the points equidistant from the points A(-2,3) & B(6,-5) is

Find the equation to the locus of a point equidistant from the points A(1,3) and B(-2,1) .

Determined the equation to the locus of the point which is equidistant from the points ( 2 , - 2 , - 4) and ( - 3 , 1 , 2) .

The equation of the locus of the points equidistant from the point A(-2,3) and B(6,-5) is

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

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

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: