Let the vertices of a triangle are A(8,6)B(8,−2) and C(2,−2)
Now, ‌‌AB=√(8−8)2+(6+2)2 =√0+82=8
BC=√(2−8)2+(−2+2)2=√36+0=6
and CA=√(8−2)2+(6+2)2 =√62+82 =√36+64=√100=10 Now, ‌‌AB2+BC2=(8)2+(6)2 =64+36=100=AC2 So, ABC is a right angled triangle and right angled at B. We know that, in a right angled triangle, the circumcenter is the mid-point of hypotenuse. ∴ Mid-point of AC=(‌