Let the coordinates of the third corner of the isosceles triangle be #(x,y)#. This point is equidistant from other two corners.
So
#(x-1)^2+(y-7)^2=(x-5)^2+(y-3)^2#
#=>x^2-2x+1+y^2-14y+49=x^2-10x+25+y^2-6y+9#
#=>8x-8y=-16#
#=>x-y=-2#
#=>y=x+2#
Now the perpendicular drawn from #(x,y)# on the line segment joining two given corners of triangle will bisect the side and the coordinates of this mid point will be #(3,5)#.
So height of the triangle
#H=sqrt((x-3)^2+(y-5)^2)#
And base of the triangle
#B=sqrt((1-5)^2+(7-3)^2)=4sqrt2#
Area of the triangle
#1/2xxBxxH=6#
#=>H=12/B=12/(4sqrt2)#
#=>H^2=9/2#
#=>(x-3)^2+(y-5)^2=9/2#
#=>(x-3)^2+(x+2-5)^2=9/2#
#=>2(x-3)^2=9/2#
#=>(x-3)^2=9/4#
#=>x=3/2+3=9/2=4.5#
So #y=x+2=4.5+2=6.5#
Hence length of each equal sides
#=sqrt((5-4.5)^2+(3-6.5)^2)#
#=sqrt(0.25+12.25)=sqrt12.5=2.5sqrt2#
Hence lengths of three sides are #2.5sqrt2,2.5sqrt2,4sqrt2#
