There are 22 Methods to deal with the Problem :
Method I :-
Let P(x,y)P(x,y) be equidistant from the pts. A(-1,3) and B(3.-1)A(−1,3)andB(3.−1).
Then, by what is given, PA^2=PB^2PA2=PB2
:. (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. Mof 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.!
