Question

已知三角形ABC顶点 证明 ABC是直角三角形

Answer 1

AB^2=(-4-2)^2+(6-2)^2=36+16=52

BC^2=(2-2)^2+(2-6)^2=16

AC^2=(-4-2)^2+(6-6)^2=36

AB^2=52=16+36=BC^2+AC^2

所以△ABC是直角三角形