還沒學定點分點公式就這樣:
DP=λPC ==> DA+AP=λ(PA+AC) ==> (1+λ)AP=AD+λAC
所以有AP=(xa+λb)/(1+λ)
下面就一樣了.相當于推導一次公式.
設DP=λPC,BP=μPE,由定比分點公式得
AP=(xa+λb)/(1+λ)=(a+μyb)/(1+μ)
因為a,b不共線,所以
x/(1+λ)=1/(1+μ),
λ/(1+λ)=μy/(1+μ),
解得
λ=(xy-y)/(y-1),μ=(x-1)/(xy-x)
代入得
AP=[(xy-x)a+(xy-y)b]/(xy-1)