题目内容
已知直线l经过直线2x+y-5=0与x-2y=0的交点.
(1)点A(5,0)到l的距离为3,求l的方程;
(2)求点A(5,0)到l的距离的最大值.
(1)点A(5,0)到l的距离为3,求l的方程;
(2)求点A(5,0)到l的距离的最大值.
(1)x=2或4x-3y-5=0
(2)
(2)

解:(1)经过两已知直线交点的直线系方程为(2x+y-5)+λ(x-2y)=0,
即(2+λ)x+(1-2λ)y-5=0.
∴
=3.
即2λ2-5λ+2=0,
∴λ=2或
.
∴l的方程为x=2或4x-3y-5=0.
(2)由

解得交点P(2,1),如图,过P作任一直线l,设d为点A到l的距离,则d≤|PA|(当l⊥PA时等号成立).
∴dmax=|PA|=
.
即(2+λ)x+(1-2λ)y-5=0.
∴

即2λ2-5λ+2=0,
∴λ=2或

∴l的方程为x=2或4x-3y-5=0.
(2)由


解得交点P(2,1),如图,过P作任一直线l,设d为点A到l的距离,则d≤|PA|(当l⊥PA时等号成立).
∴dmax=|PA|=


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