向量AB=(4,3) 向量BC=(m,-2) 若A,B,C三點可構(gòu)成等腰三角形,m=?

數(shù)學(xué)人氣：892 ℃時間：2019-08-18 20:17:15

優(yōu)質(zhì)解答

已知:AB = (4,3),BC = (m,-2)
則:AC = AB + BC = (4,3) + (m,-2) = (4 + m,1)
對於A,B,C三點可構(gòu)成等腰三角形,有以下情況:
(1) 以點A為頂點,則∣AB∣=∣AC∣
得√(4² + 3²) = √[(4+m)² + 1²]
25 = (4+m)² + 1
化簡得:m = -4 ± 2√6
(2) 以點B為頂點,則∣AB∣=∣BC∣
得√(4² + 3²) = √[m² + (-2)²]
25 = m² + 4
化簡得:m = ± √21
(3) 以點C為頂點,則∣AC∣=∣BC∣
得√[(4+m)² + 1²] = √[m² + (-2)²]
m² + 8m + 16 + 1 = m² + 4
化簡得:m = -13/8