2019屆高中數(shù)學(xué)第三章直線與方程3.3.2兩點(diǎn)間的距離課后篇鞏固探究含解析新人教A版必修2.docx

3.3.2兩點(diǎn)間的距離課后篇鞏固提升1.已知點(diǎn)A(-2,-1),B(a,3),且|AB|=5,則a的值為()A.1B.-5C.1或-5D.-1或5解析由|AB|=(a+2)2+(3+1)2=5,得(a+2)2=9,解得a=1或-5.答案C2.已知ABC的頂點(diǎn)A(2,3),B(-1,0),C(2,0),則ABC的周長是()A.23B.3+23C.6+32D.6+10解析|AB|=(-1-2)2+(0-3)2=32,|BC|=3,|AC|=(2-2)2+(0-3)2=3,ABC的周長為6+32.答案C3.已知點(diǎn)A(x,5)關(guān)于點(diǎn)(1,y)的對(duì)稱點(diǎn)為(-2,-3),則點(diǎn)P(x,y)到原點(diǎn)的距離是()A.2B.4C.5D.17解析根據(jù)中點(diǎn)坐標(biāo)公式,得x-22=1,且5-32=y.解得x=4,y=1,所以點(diǎn)P的坐標(biāo)為(4,1),則點(diǎn)P(x,y)到原點(diǎn)的距離d=(4-0)2+(1-0)2=17.答案D4.到點(diǎn)A(1,3),B(-5,1)的距離相等的動(dòng)點(diǎn)P滿足的方程是()A.3x-y-8=0B.3x+y+4=0C.3x-y+6=0D.3x+y+2=0解析設(shè)P(x,y),則(x-1)2+(y-3)2=(x+5)2+(y-1)2,即3x+y+4=0.答案B5.若點(diǎn)P(x,y)在直線4x+3y=0上,且x,y滿足-14x-y7,則點(diǎn)P到坐標(biāo)原點(diǎn)距離的取值范圍是()A.0,5B.0,10C.5,10D.5,15解析點(diǎn)P(x,y)在直線4x+3y=0上,且x,y滿足-14x-y7,-6x3.線段4x+3y=0(-6x3)過原點(diǎn),點(diǎn)P到坐標(biāo)原點(diǎn)的最近距離為0.又點(diǎn)(-6,8)在線段上,點(diǎn)P到坐標(biāo)原點(diǎn)的最遠(yuǎn)距離為(-6)2+82=10.點(diǎn)P到坐標(biāo)原點(diǎn)距離的取值范圍是0,10.答案B6.光線從點(diǎn)A(-3,5)射到x軸上,經(jīng)x軸反射后經(jīng)過點(diǎn)B(2,10),則光線從A到B的距離為()A.52B.25C.510D.105解析點(diǎn)B(2,10)關(guān)于x軸的對(duì)稱點(diǎn)為B(2,-10),由對(duì)稱性可得光線從A到B的距離為|AB|=(-3-2)2+5-(-10)2=510.選C.答案C7.若動(dòng)點(diǎn)P的坐標(biāo)為(x,1-x),xR,則動(dòng)點(diǎn)P到原點(diǎn)的最小值是.解析|PO|=x2+(1-x)2=2x2-2x+1=2x-122+1222.答案228.已知M(1,0),N(-1,0),點(diǎn)P在直線2x-y-1=0上移動(dòng),則|PM|2+|PN|2的最小值為.解析點(diǎn)P在直線2x-y-1=0上,可設(shè)P的坐標(biāo)為(a,2a-1),|PM|2+|PN|2=(a-1)2+(2a-1)2+(a+1)2+(2a-1)2=10a2-8a+4=10a-252+125.|PM|2+|PN|2的最小值為125.答案1259.兩直線3ax-y-2=0和(2a-1)x+5ay-1=0分別過定點(diǎn)A,B,則|AB|的值為.解析直線3ax-y-2=0過定點(diǎn)A(0,-2),直線(2a-1)x+5ay-1=0過定點(diǎn)B-1,25,由兩點(diǎn)間的距離公式,得|AB|=135.答案13510.已知正方形ABCD的邊長為4,若E是BC的中點(diǎn),F是CD的中點(diǎn),試建立坐標(biāo)系,求證:BFAE.證明建立平面直角坐標(biāo)系,如圖所示,則B(4,0),E(4,2),F(2,4),A(0,0).設(shè)直線AE,BF的斜率分別為kAE,kBF,則kAE=24=12,kBF=4-02-4=-2.于是kAEkBF=12(-2)=-1,故BFAE.11.(選做題)在x軸上求一點(diǎn)P,使得(1)P到A(4,1)和B(0,4)的距離之差最大,并求出最大值;(2)P到A(4,1)和C(3,4)的距離之和最小,并求出最小值.解(1)如圖,設(shè)直線BA與x軸交于點(diǎn)P,此時(shí)P為所求點(diǎn),且|PB|-|PA|=|AB|=(0-4)2+(4-1)2=5.直線BA的斜率kBA=1-44=-34,直線BA的方程為y=-34x+4.令y=0,得x=163,即P163,0.故距離之差最大值為5,此時(shí)P點(diǎn)的坐標(biāo)為163,0.(2)作A關(guān)于x軸的對(duì)稱點(diǎn)A,則A(4,-1),連接CA,則|CA|為所求最小值,直線CA與x軸交點(diǎn)為所求點(diǎn).又|CA|=(4-3)2+(-1-4)2=2