Let A(3, 4), B(-2, 3) and C(x,y) be the three vertices of the equilateral triangle then, AB2=BC2=CA2
Now, AB2=BC2
⇒ x2+y2+4x-6y+13=26
⇒ x2+y2+4x-6y-13=0 .......(i)
AB2=CA2
⇒ 26-(x2+y2-6x-8y+25)=0
⇒ x2+y2-6x-8y-1=0 ........(ii)
Subtracting (ii) from (i) we get,
10x+2y-12=0
⇒ 5x+y=6 ......(iii)
⇒ 5x=6-y
⇒ x=(6-y)/5
Subtracting x=(6-y)/5 in (i) we get
⇒ 26y2-32y+6-325=0
⇒ 26y2-32y-319=0
D=b2-4ac
D=(-32)2-4x26x(-319)
=1024+33176
=34200
Therefore, y=(32+185)/52=217/52 or y=(32-185)/52= -153/52
Substituting y=217/52 in (iii)
5x+217/52=6
5x=6-217/52=95/52
x=19/52
Again substituting y=-153/52 in (iii)
5x-153/52=6
5x=6+153/52=465/52
x=93/52
Therefore, the coordinates of the third vertex are (19/52,217/52) or (93/52,-153/52)