The equation of the locus of points equidistant from `(-1-1)` and `(4,2)` is

Find the equation of the set of points equidistant from (-1, -1) and (4, 2)

The locus of the point equidistant from (1,-1) and (-1,1) is

Locus of the points equidistant from the point (-1, -1) and (4, 2) is

The equation to the locus of points equidistant from the points (-3, 4),(3, 4) is

The equation to the locus of points equidistant from the points ( 2 , 3 ) , ( - 2, 5 ) is

The equation of locus of a point equidistant from A(2 0) and the y -axis is/are

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_2+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_2+b_2)y+c=0 , then the value of C is