点P(0,1)在直线ax+y-b=0上的射影是点Q(1,0).则直线ax-y+b=0关于直线x+y-1=0对称的直线方程是_______.
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解:过PQ两点直线的斜率为:(1-0)/(0-1)=-1直线ax+y-b=0的斜率为:-a由已知两直线垂直,可得:a=-1.又∵直线ax+y-b=0过点Q(1,0).b=-1则直线ax-y+b=0化为:-x-y-1=0,即:x+y+1=0,它与直线x+y-1=0平行.∴设其对称的直线方程为:x+y+m=0则d=|m-(-1)|/√2=|1-(-1)|/=2/√2即:m+1=±2m=-3或m=1(舍)则所求直线方程是x+y-3=0