試求點P(0,0)到直線L:2x+y-5=0的距離?
回答 (2)
2017-10-12 6:32 am
Sol
d=｜2*0+0－5｜/√(2^2+1^2)
=5/√5
=√5
or
2x+y－5=0
y=5－2x
A(x，y)為L上一點
PA^2=(x－0)^2+(y－0)^2
=x^2+y^2
=x^2+(5－2x)^2
=x^2+(4x^2－20x+25)
=5x^2－20x+25
=5(x^2－4x)+25
=5(x^2－4x+4)+5
=5(x－2)^2+5
So
PA^2>=5
PA>=√5
d=｜2*0+0－5｜/√(2^2+1^2)
=5/√5
=√5
or
2x+y－5=0
y=5－2x
A(x，y)為L上一點
PA^2=(x－0)^2+(y－0)^2
=x^2+y^2
=x^2+(5－2x)^2
=x^2+(4x^2－20x+25)
=5x^2－20x+25
=5(x^2－4x)+25
=5(x^2－4x+4)+5
=5(x－2)^2+5
So
PA^2>=5
PA>=√5
2017-10-12 5:10 pm
5
收錄日期: 2021-04-30 22:30:50
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20171011180706AAQpiXx