A(m,n)代入到中线方程中有2m-3n+6=0
BC的斜率K＝（3－1）/(-2-2)=-1/2,故BC的方程是y-1=-1/2(x-2)
即有x+2y-4=0
A到BC的距离d=|m+2n-4|/根号（1+4）＝｜m+2n-4|/根号5
又｜BC|=根号[（2+2）^2+(1-3)^2]=2根号5
故面积S＝1/2|BC|*d=1/2*2根号5*｜m+2n-4|/根号5＝7
即有|m+2n-4|=7
即有m+2n-4=7或m+2n-4=-7
又有2m-3n+6=0
解得m=3,n=4或m=-3,n=0