可设直线方程为y-1=k(x-2) (k≠0,否则与椭圆相切),设两交点分别为（x1,y1）,(x2,y2),则x1^2/2+y1^2=1,x2^2/2+y2^2=1,两式相减得(x1+x2)(x1-x2)/2+(y1+y2)(y1-y2)=0,显然x1≠x2(两点不重合),故（x1+x2）/2+(y1+y2)(y1...