向量AB=(4,3) 向量BC=(m,-2) 若A,B,C三点可构成等腰三角形,m=?

已知:AB = (4,3),BC = (m,-2)
则:AC = AB + BC = (4,3) + (m,-2) = (4 + m,1)
对於A,B,C三点可构成等腰三角形,有以下情况:
(1) 以点A为顶点,则∣AB∣=∣AC∣
得√(4² + 3²) = √[(4+m)² + 1²]
25 = (4+m)² + 1
化简得:m = -4 ± 2√6
(2) 以点B为顶点,则∣AB∣=∣BC∣
得√(4² + 3²) = √[m² + (-2)²]
25 = m² + 4
化简得:m = ± √21
(3) 以点C为顶点,则∣AC∣=∣BC∣
得√[(4+m)² + 1²] = √[m² + (-2)²]
m² + 8m + 16 + 1 = m² + 4
化简得:m = -13/8