两点式方程如何化为截距式方程
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/13 09:32:55
两点式方程如何化为截距式方程
![两点式方程如何化为截距式方程](/uploads/image/z/18477950-14-0.jpg?t=%E4%B8%A4%E7%82%B9%E5%BC%8F%E6%96%B9%E7%A8%8B%E5%A6%82%E4%BD%95%E5%8C%96%E4%B8%BA%E6%88%AA%E8%B7%9D%E5%BC%8F%E6%96%B9%E7%A8%8B)
(y-y1)/(x-x1)=(y2-y1)/(x2-x1) (直线两点式方程)
(y-y1)(x2-x1)=(x-x1)(y2-y1)
(x2-x1)y-(x2-x1)y1=(y2-y1)x-(y2-y1)x1
(y1-y2)x+(x2-x1)y=(x2-x1)y1-(y2-y1)x1
(y1-y2)x+(x2-x1)y=x2y1-x1y2
(y1-y2)x/(x2y1-x1y2)+(x2-x1)y/(x2y1-x1y2)=1
x/[(x2y1-x1y2)/(y1-y2)]+y/[(x2y1-x1y2)/(x2-x1)]=1(直线的截距式)
(y-y1)(x2-x1)=(x-x1)(y2-y1)
(x2-x1)y-(x2-x1)y1=(y2-y1)x-(y2-y1)x1
(y1-y2)x+(x2-x1)y=(x2-x1)y1-(y2-y1)x1
(y1-y2)x+(x2-x1)y=x2y1-x1y2
(y1-y2)x/(x2y1-x1y2)+(x2-x1)y/(x2y1-x1y2)=1
x/[(x2y1-x1y2)/(y1-y2)]+y/[(x2y1-x1y2)/(x2-x1)]=1(直线的截距式)