Given A (-3,3), B (1,11) and C (3,15)
AB = sqrt[(3 - 11)^2 + (-3 - 1)^2] = sqrt[8^2 + 4^2] = sqrt[64 + 16] = sqrt[80] = 4sqrt(5)
BC = sqrt[(11 - 15)^2 + (1 - 3)^2] = sqrt[4^2 + 2^2] = sqrt[16 + 4] = sqrt[20] = 2sqrt[5]
Therefore AB + BC = 6sqrt[5]
AC = sqrt[(3 - 15)^2 + (-3 - 3)^2] = sqrt[12^2 + 6^2] = sqrt[144 + 36] = sqrt[180] = 6sqrt[5]
AC = 6sqrt[5]
Since AC = AB + BC, then B lies on the straight line AC