直线与曲线交点求解问题
 

最近做化工原理课程设计，苯－氯苯的气液平衡数据x-y图和q线方程为：q=-2*x+2.184编程作图如下：
>> x=[0,0.0115,0.0454,0.0879,0.131,0.185,0.276,0.44,0.543,0.678,0.818,1];
y=[0,0.0566,0.201,0.343,0.456,0.563,0.665,0.782,0.847,0.911,0.957,1;];
plot(x,y);hold on ;
fplot('-2*x+2.184',[0.6,0.8],1e-4);
hold off;
hold on;fplot('x',[0 1],1e-4);hold off;
text(x,y,['(' num2str(x) ',' num2str(y) ')'])
本想求得直线与曲线交点，但结果看不懂。希望高手给予小菜指导，在此谢过！
PS：图形与交点坐标都要。我的图如下[IMG]http://winsun815.blog.163.com/editPhoto.do?albumId=-1&photoId=fks_087070093086080067087094086066072085084066082082095067087[/IMG]

回复: 直线与曲线交点求解问题
 

我用CAD求解的结果是（ 0.664，0.895）

回复: 直线与曲线交点求解问题
 

先拟合数据得到函数，然后求交点。。