There are #2# Methods to deal with the Problem :
Method I :-
Let #P(x,y)# be equidistant from the pts. #A(-1,3) and B(3.-1)#.
Then, by what is given, #PA^2=PB^2#
#:. (x+1)^2+(y-3)^2=(x-3)^2+(y+1)^2#.
#:. (x+1)^2-(x-3)^2+(y-3)^2-(y+1)^2=0#.
#:. (x+1+x-3)(x+1-x+3)+(y-3+y+1)(y-3-y-1)=0#.
#:. 4(2x-2)-4(2y-2)=0#.
#:. x-1-y+1=0#.
#:. x-y=0#, is the desired eqn. of the locus.
Method II :-
We know, from Geometry, that, the locus of the pts. equidistant from two fixed pts. in the plane is the #bot#-bisector of the segment joining those two fixed pts.
So, to find the eqn. of the locus, we have to find the eqn. of #bot#-bisector, say, #l#, of segment #AB#.
The mid-pt. #M#of seg. #AB# is #M((-1+3)/2,(3-1)/2)=M(1,1)#
Slope of #AB# is #(3+1)/(-1-3)=-1#
Since, #l bot "line" AB#. the slope of #l# is #1#.
Thus, the slope of #l# is #1#, and, #M in l#, using Slope-Pt. Form,
the eqn. of "l# is : y-1=1(x-1), i.e., x-y=0#, as before!
Enjoy Maths.!
