直线方程两点式化一般式、截距式的化法?

如题所述

第1个回答  2020-05-30
(y-y2)/(y1-y2)=(x-x2)/(x1-x2)
(x1-x2)y-y2(x1-x2)=(y1-y2)x-x2(y1-y2)
(y1-y2)x-(x1-x2)y+(x1y2-x2y2-x2y1+x2y2)=0
截距式
(y1-y2)x-(x1-x2)y=(x2y2+x2y1-x1y2-x2y2)
x/[(x2y2+x2y1-x1y2-x2y2)/y1-y2)]+y/[(x1y2-x2y2-x2y1+x2y2)/(x1-x2)]=1
相似回答