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定理:对于两点 P_1(x_1,y_1)和 P_2(x_2,y_2),若直线Ax+By+C=0与直线 P_1P_2交于 P_0(x_0,y_0),而(P_1P_0)/(P_0P_2)=λ(λ≠-1),则λ=(Ax_1+By_1+C)/(Ax_2+By_2+C)证明:因 P_1P_0:P_0P_2=λ,故x_0=(x_1+λx_2)/(1+λ),y_0=(y_1+λy_2)/(1+λ).因 P_0在直线 Ax+By+C=0上,故A((x_1+λx_2)/(1+λ))+B((y_1+λy_2)/(1+λ))+C=0.解之得:λ=(Ax_1+By_1+C)/(Ax_2+By_2+C)下面举例说明上面定理的应用例1 已知点 A(-20)R(22)和 C(0.5)过
Theorem: For two points P_1(x_1,y_1) and P_2(x_2,y_2), if the straight line Ax+By+C=0 crosses P_(x_0,y_0) with the straight line P_1P_2, and (P_1P_0)/(P_0P_2)=λ (λ≠-1), then λ=(Ax_1+By_1+C)/(Ax_2+By_2+C) Proof: Since P_1P_0:P_0P_2=λ, x_0=(x_1+λx_2)/(1+λ), y_0 =(y_1+λy_2)/(1+λ). Since P_0 is on the straight line Ax+By+C=0, A((x_1+λx_2)/(1+λ))+B((y_1+λy_2)/ (1+λ))+C=0. The solution is: λ=(Ax_1+By_1+C)/(Ax_2+By_2+C) The following is an example of application of the above theorem. 1 Known point A(-20)R (22) and C (0.5)