Name the isosceles right-triangle as #DeltaABC#, and let
#AC# be the hypotenuse, with #A=A(1,3) and C=(-4,1)#.
Consequently, #BA=BC#.
So, if #B=B(x,y)#, then, using the distance formula,
#BA^2=BC^2rArr(x-1)^2+(y-3)^2=(x+4)^2+(y-1)^2#.
#rArrx^2-2x+1+y^2-6y+9=x^2+8x+16+y^2-2y+1#
#rArr10x+4y+7=0.........................................................<<1>>#.
Also, as #BAbotBC," slope of "BAxx"slope of "BC=-1#.
#:.{(y-3)/(x-1)}{(y-1)/(x+4)}=-1#.
#:.(y^2-4y+3)+(x^2+3x-4)=0#.
#:.x^2+y^2+3x-4y-1=0..............................<<2>>#.
#<<1>> rArr y=-(10x+7)/4...<<1'>>#. Sub.ing in #<<2>>#, we get,
#x^2+(-(10x+7)/4)^2+3x-4(-(10x+7)/4)-1=0#.
#:.16x^2+(100x^2+140x+49)+48x+160x+112-16=0#
#:. 116x^2+348x+145=0#.
#"Dividing by "29," we have, "4x^2+12x+5=0, or, #
#4x^2+12x=-5#,
#rArr4x^2+12x+9=-5+9......[because," completing square]"#,
#rArr (2x+3)^2=4=2^2 :. 2x+3=+-2 :. 2x=-3+-2#.
#:. x=-1/2, or, x=-5/2#.
#<<1'>> rArr y=-1/2, or, y=9/2#.
Hence, the remaining vertex of the triangle can be, either
#(-1/2,-1/2), or, (-5/2,9/2)#.
