u02115
幼苗
共回答了11个问题采纳率:81.8% 举报
设直线l方程:y-1=k(x-1)y=kx-k+1当y=0,0=kx-k+1x=(k-1)/k,x轴截距当x=0,y=-k+1,y轴截距(1)s=f(k)=1/2*(k-1)/k*(-k+1)=(k-1)(1-k)/(2k)(2)s=(k-1)(1-k)/(2k)=(k-1-k+k)/(2k)=(-k+2k-1)/(2k)ds/dk=1/(4k)*[(-2k+2)(2k)-(-k+2k-1)(2)]=(1-k)/(2k)令ds/dk=01-k=0k=1或-1,(直线l经过正x,y轴,斜率k必为负数,所以舍去k=1)而ds/dk=-1/kds/dk|(k=-1)=1>0s的最小值为f(-1)=(-1-1)[1-(-1)]/[2(-1)]=-2*2/(-2)=2相应直线l的方程为y=(-1)x-(-1)+1y=-x+2
1年前
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