The given points are A(0, 4, 1), B(2, 3, −1) and C(4, 5, 0)
AB2 = (0 − 2)2 + (4 − 3)2 + (1 + 1)2 = 4 + 1 + 4 = 9
BC2 = (2 − 4)2 + (3 − 5)2 + (−1 − 0)2 = 4 + 4 + 1 = 9
CA2= (4 − 0)2+ (5 − 4)2+ (0 − 1)2 = 16 + 1 + 1 = 18
∴ AB2 + BC2 = CA2
Hence ΔABC is a right-angled triangle.
