已知两点M(-2,2),N(5,
已知两点M(-2,2),N(5,-2)在坐标轴上求一点P,使∠MPN为直角,并求出线段MN的垂直平分线
①当点P在x轴上时,不妨设点P(a,0)
则:
PM^2=(a+2)^2+4=a^2+4a+8
PN^2=(a-5)^2+4=a^2-10a+29
MN^2=(5+2)^2+(-2-2)^2=49+16=65
因为∠MPN为直角,所以由勾股定理有:PM^2+PN^2=MN^2
即:a^2+4a+8+a^2-10a+29=65
===> 2a^2-6a-28=0
===> a^2-3a-14=0
===> a=[3±√(9+56)]/2=(3±√65)/2
则,点P((3±√65)/2,0...全部
已知两点M(-2,2),N(5,-2)在坐标轴上求一点P,使∠MPN为直角,并求出线段MN的垂直平分线
①当点P在x轴上时,不妨设点P(a,0)
则:
PM^2=(a+2)^2+4=a^2+4a+8
PN^2=(a-5)^2+4=a^2-10a+29
MN^2=(5+2)^2+(-2-2)^2=49+16=65
因为∠MPN为直角,所以由勾股定理有:PM^2+PN^2=MN^2
即:a^2+4a+8+a^2-10a+29=65
===> 2a^2-6a-28=0
===> a^2-3a-14=0
===> a=[3±√(9+56)]/2=(3±√65)/2
则,点P((3±√65)/2,0)
②当点P在y轴上时,不妨设点P(0,b)
则:
PM^2=4+(b-2)^2=b^2-4b+8
PN^2=25+(b+2)^2=b^2+4b+29
MN^2=(5+2)^2+(-2-2)^2=49+16=65
因为∠MPN为直角,所以由勾股定理有:PM^2+PN^2=MN^2
即:b^2-4b+8+b^2+4b+29=65
===> 2b^2=28
===> b^2=14
===> b=±√14
则,点P(0,±√14)
线段MN的中点为x=(-2+5)/2=3/2,y=(2-2)/2=0
即,中点为(3/2,0)
又MN所在直线的斜率为k=(-2-2)/(5+2)=-4/7
所以,其垂直平分线的斜率为k'=7/4
所以,其垂直平分线为:y-0=(7/4)*[x-(3/2)]
===> y=(7/4)x-(21/8)
===> 8y=14x-21
===> 14x-8y-21=0。
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