设 点P（5,-2）关于直线2x+y-1=0的对称点P'(x',y')则直线l 2x+y-1=0为线段PP'的垂直平分线∴PP'的中点M（（5+x')/2,(-2+y')/2)在l上且kPP'*(-2)=-1∴(y'+2)/(x'-5)=1/22(5+x')/2+(y'-2)/2-1=0解得;x'=-3/5 y'=-24/5...明白了｛(y'+2)/(x'-5)=1/2｛ 2(5+x')/2+(y'-2)/2-1=0