Question

直线与曲线的交点个数直线(c-d)(x-b)-(a-b)(y-d)=0与曲线(x-a)(x-b)+(y-c)(y-d)=

Answer 1

将直线转化为x=b+(a-b)(y-d)/(c-d) c-d≠0

代入曲线方程（a-b）[b-a+(a-b)(y-d)/(c-d)]+(c-d)(y-d)(y-c)=0

两个交点