第2章平面解析幾何初步（培優(yōu)課對(duì)稱問(wèn)題的解法）課件高二上學(xué)期數(shù)學(xué)選擇性

培優(yōu)課對(duì)稱問(wèn)題的解法第2章平面解析幾何初步湘教版

數(shù)學(xué)

選擇性必修第一冊(cè)課標(biāo)要求1.掌握平面上點(diǎn)關(guān)于點(diǎn)、線的對(duì)稱問(wèn)題的解法;2.能夠根據(jù)對(duì)稱思想,求解實(shí)際問(wèn)題.重難探究·能力素養(yǎng)速提升目錄索引

學(xué)以致用·隨堂檢測(cè)促達(dá)標(biāo)重難探究·能力素養(yǎng)速提升探究點(diǎn)一點(diǎn)關(guān)于點(diǎn)、線的對(duì)稱問(wèn)題【例1】

直線l:2x+3y-6=0關(guān)于點(diǎn)(-1,2)對(duì)稱的直線方程是(

)A.3x-2y-10=0B.3x-2y-23=0C.2x+3y-4=0D.2x+3y-2=0D解析

(方法1)設(shè)所求對(duì)稱直線方程上的點(diǎn)的坐標(biāo)為(x,y),則其關(guān)于點(diǎn)(-1,2)對(duì)稱的對(duì)稱點(diǎn)的坐標(biāo)為(-2-x,4-y).因?yàn)辄c(diǎn)(-2-x,4-y)在直線l:2x+3y-6=0上,所以2(-2-x)+3(4-y)-6=0,整理得2x+3y-2=0.故選D.(方法2)設(shè)直線l:2x+3y-6=0關(guān)于點(diǎn)(-1,2)對(duì)稱的直線為l1,則l∥l1.設(shè)直線l1的方程為2x+3y+t=0(t≠-6).在直線2x+3y-6=0上任取一點(diǎn)P(0,2),點(diǎn)P(0,2)關(guān)于(-1,2)的對(duì)稱點(diǎn)為P'(-2,2),將P'(-2,2)代入2x+3y+t=0(t≠-6)得t=-2.故所求方程為2x+3y-2=0.故選D.規(guī)律方法

1.點(diǎn)關(guān)于點(diǎn)成中心對(duì)稱的解法若兩點(diǎn)P1(x1,y1),P2(x2,y2)關(guān)于點(diǎn)P0(x0,y0)中心對(duì)稱,則2.線關(guān)于點(diǎn)成中心對(duì)稱問(wèn)題的解法(1)設(shè)出所求直線上的點(diǎn)的坐標(biāo)P(x,y),求出點(diǎn)P(x,y)關(guān)于M的對(duì)稱點(diǎn)P'的坐標(biāo),將點(diǎn)P'的坐標(biāo)代入已知直線,所得方程即為所求直線的方程;(2)根據(jù)直線l關(guān)于直線外一點(diǎn)M的對(duì)稱直線與已知直線互相平行可得對(duì)稱直線l'的斜率,再取直線l上任意一特殊點(diǎn),求其關(guān)于點(diǎn)M的對(duì)稱點(diǎn),將該點(diǎn)代入直線l'中,即可求出直線l'的方程.變式訓(xùn)練1已知直線l1與l2關(guān)于原點(diǎn)對(duì)稱,若l1的方程是x+2y-3=0,則l2的方程是(

)A.x+2y+3=0 B.x-2y+3=0C.2x+y+3=0 D.2x-y+3=0A解析

因?yàn)橹本€l1與l2關(guān)于原點(diǎn)對(duì)稱,則只需將l1的方程中x改為-x,y改為-y,可得l2的方程是-x+2(-y)-3=0,即x+2y+3=0,故選A.探究點(diǎn)二軸對(duì)稱問(wèn)題角度1點(diǎn)關(guān)于直線的對(duì)稱問(wèn)題【例2】

求點(diǎn)A(-2,3)關(guān)于直線3x-y-1=0對(duì)稱的點(diǎn)A'的坐標(biāo).分析

設(shè)A(-2,3)關(guān)于直線的對(duì)稱點(diǎn)為A'(x0,y0),根據(jù)對(duì)稱性的特征建立方程組求解.

解

設(shè)A'(x0,y0),由題意,所以點(diǎn)A關(guān)于直線3x-y-1=0對(duì)稱的點(diǎn)A'為(4,1).規(guī)律方法

1.點(diǎn)關(guān)于直線對(duì)稱問(wèn)題的解法若兩點(diǎn)P1(x1,y1),P2(x2,y2)關(guān)于直線l:Ax+By+C=0(A,B不同時(shí)為0)對(duì)稱,則線段P1P2的中點(diǎn)在直線l上,且直線P1P2⊥直線l,2.點(diǎn)關(guān)于特殊的直線的對(duì)稱問(wèn)題的結(jié)論

點(diǎn)的坐標(biāo)對(duì)稱直線結(jié)論點(diǎn)P(x0,y0)y=x(y0,x0)y=-x(-y0,-x0)x+y+t=0(-t-y0,-t-x0)x-y+t=0(y0-t,x0+t)變式訓(xùn)練2[2024甘肅高二期末](多選題)已知點(diǎn)M(1,0)關(guān)于直線mx-y+1=0(m∈R)的對(duì)稱點(diǎn)N在直線x+y=0上,則實(shí)數(shù)m的值為(

)AC解析

因?yàn)辄c(diǎn)N在直線x+y=0上,設(shè)點(diǎn)N(a,-a),又M(1,0),化簡(jiǎn)為m(a+1)+a+2=0.①因?yàn)镸N與直線mx-y+1=0垂直,角度2線關(guān)于線成軸對(duì)稱問(wèn)題【例3】

已知直線l:x-y-1=0,l1:x-y+3=0,l2:2x-y-1=0.(1)求直線l1關(guān)于直線l的對(duì)稱直線l1'的方程;(2)求直線l2關(guān)于直線l的對(duì)稱直線l2'的方程.解

(1)因?yàn)閘1∥l,所以l1'∥l.設(shè)直線l1'的方程為x-y+c=0(c≠3,且c≠-1).在直線l1上取點(diǎn)M(0,3),設(shè)點(diǎn)M關(guān)于直線l的對(duì)稱點(diǎn)為M'(a,b),把點(diǎn)M'的坐標(biāo)代入直線l1'的方程,得4-(-1)+c=0,解得c=-5.所以直線l1'的方程為x-y-5=0.規(guī)律方法

直線關(guān)于直線的對(duì)稱直線方程的求解方法設(shè)直線l1關(guān)于直線l的對(duì)稱直線為l2.直線l1與直線l的關(guān)系方法過(guò)程平行l(wèi)1∥l2∥l,取直線l1上一點(diǎn),求該點(diǎn)關(guān)于直線l的對(duì)稱點(diǎn)利用兩直線平行及點(diǎn)關(guān)于直線對(duì)稱求解相交求l與l1的交點(diǎn),再在直線l1上取一點(diǎn)(不是交點(diǎn)),求該點(diǎn)關(guān)于直線l的對(duì)稱點(diǎn)利用點(diǎn)關(guān)于直線對(duì)稱求方程變式訓(xùn)練3(1)求直線l1:2x-y+6=0關(guān)于直線l:2x-y+1=0對(duì)稱的直線l2的方程;(2)求直線2x+y-4=0關(guān)于直線x-y+1=0對(duì)稱的直線的方程.解

(1)設(shè)直線l2的方程為2x-y+c=0(c≠6且c≠1),在直線l1:2x-y+6=0上取點(diǎn)P(-2,2),設(shè)點(diǎn)P關(guān)于直線l:2x-y+1=0的對(duì)稱點(diǎn)為P'(m,n),則(2)因?yàn)橹本€2x+y-4=0與直線x-y+1=0相交,聯(lián)立直線方程可得

探究點(diǎn)三點(diǎn)關(guān)于直線對(duì)稱問(wèn)題的應(yīng)用【例4】

在直線l:3x-y-1=0上,求點(diǎn)P和Q,使得(1)點(diǎn)P到點(diǎn)A(4,1)和B(0,4)的距離之差最大;(2)點(diǎn)Q到點(diǎn)A(4,1)和C(3,4)的距離之和最小.解

(1)如圖所示,設(shè)點(diǎn)B關(guān)于直線l的對(duì)稱點(diǎn)B'的坐標(biāo)為(a,b),聯(lián)立①②可得a=3,b=3,即B'點(diǎn)的坐標(biāo)為(3,3).即直線l與直線AB'的交點(diǎn)坐標(biāo)為P(2,5),且此時(shí)點(diǎn)P到點(diǎn)A,B的距離之差最大.(2)如圖所示,設(shè)點(diǎn)C關(guān)于直線l的對(duì)稱點(diǎn)為C'(m,n),規(guī)律方法

求直線上一點(diǎn)到兩定點(diǎn)的和(差)距離最值問(wèn)題的求法

兩定點(diǎn)A,B與定直線l的關(guān)系直線上尋找一點(diǎn)到兩定點(diǎn)的和最小直線上尋找一點(diǎn)到兩定點(diǎn)的差的絕對(duì)值最大兩定點(diǎn)A,B在定直線l的同側(cè)過(guò)一定點(diǎn)A(或B)關(guān)于直線l的對(duì)稱點(diǎn)A'(或B'),連接AB'(或A'B)與直線l的交點(diǎn)即為所求連接AB,則AB與直線l的交點(diǎn)即為所求兩定點(diǎn)A,B在定直線l的異側(cè)連接AB,則AB與直線l的交點(diǎn)即為所求過(guò)一定點(diǎn)A(或B)關(guān)于直線l的對(duì)稱點(diǎn)A'(或B'),連接AB'(或A'B)與直線l的交點(diǎn)即為所求變式訓(xùn)練4(1)已知A(1,4),B(8,3),點(diǎn)P在x軸上,求使|AP|+|BP|取得最小值的點(diǎn)P的坐標(biāo);(2)已知點(diǎn)A(1,3),B(5,-2),在x軸上找一點(diǎn)P使|AP|-|BP|最大,求點(diǎn)P的坐標(biāo).解

(1)A(1,4)關(guān)于x軸的對(duì)稱點(diǎn)為A'(1,-4),因?yàn)閨AP|+|BP|=|A'P|+|BP|≥|A'B|,當(dāng)A',P,B三點(diǎn)共線時(shí)取等號(hào),此時(shí)點(diǎn)P為直線A'B與x軸的交點(diǎn),直線A'B的直線方程為

,整理得y=x-5.令y=0,得x=5,所以P點(diǎn)坐標(biāo)為(5,0).(2)如圖所示,作點(diǎn)B關(guān)于x軸的對(duì)稱點(diǎn)B'(5,2),由對(duì)稱性可知|BP|=|B'P|,則|AP|-|BP|=|AP|-|B'P|.當(dāng)A,B',P三點(diǎn)不共線時(shí),由三角形三邊關(guān)系得|AP|-|B'P|<|AB'|;當(dāng)A,B',P三點(diǎn)共線時(shí),|AP|-|B'P|=|AB'|.所以|AP|-|B'P|≤|AB'|,當(dāng)且僅當(dāng)A,B',P三點(diǎn)共線時(shí),等號(hào)成立,此時(shí),直線AB'學(xué)以致用·隨堂檢測(cè)促達(dá)標(biāo)1234567891011121314151617A級(jí)必備知識(shí)基礎(chǔ)練1.點(diǎn)A(1,4)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)是(

)A.(-1,4) B.(1,-4)C.(-1,-4) D.(-4,-1)C12345678910111213141516172.點(diǎn)A(1,4)關(guān)于點(diǎn)M(0,-1)的對(duì)稱點(diǎn)坐標(biāo)是(

)A.()

B.(-1,2)C.(-1,6) D.(-1,-6)D解析

由題意可知,點(diǎn)M是所求點(diǎn)與點(diǎn)A的中點(diǎn),設(shè)所求點(diǎn)為A'(x,y),則

12345678910111213141516173.點(diǎn)A(1,4)關(guān)于x軸的對(duì)稱點(diǎn)的坐標(biāo)是(

)A.(-1,4) B.(1,-4)C.(-1,-4) D.(-4,-1)B12345678910111213141516174.點(diǎn)P(2,5)關(guān)于直線x=4的對(duì)稱點(diǎn)的坐標(biāo)是(

)A.(6,5) B.(6,-5) C.(5,6) D.(5,-6)A解析

設(shè)點(diǎn)Q(a,b)為所求的對(duì)稱點(diǎn),則由題意知b=5,且點(diǎn)Q與點(diǎn)P的中點(diǎn)在直線x=4上,因此

=4,得a=6.故所求對(duì)稱點(diǎn)是(6,5).12345678910111213141516175.點(diǎn)P(2,5)關(guān)于直線x+y=0的對(duì)稱點(diǎn)的坐標(biāo)是(

)A.(5,2) B.(-5,-2)C.(-2,-5) D.(5,-2)B解析

設(shè)對(duì)稱點(diǎn)P'(m,n),故點(diǎn)P關(guān)于直線x+y=0的對(duì)稱點(diǎn)的坐標(biāo)是(-5,-2).12345678910111213141516176.直線2x-y=2關(guān)于直線2x-y+3=0的對(duì)稱直線方程是

.

2x-y+8=0解析

直線2x-y=2可化為2x-y-2=0,則直線2x-y-2=0與直線2x-y+3=0平行.設(shè)所求直線的方程為2x-y+t=0(t≠-2,且t≠3),在直線2x-y=2上任取一點(diǎn)M(1,0),設(shè)點(diǎn)M關(guān)于直線2x-y+3=0的對(duì)稱點(diǎn)為M'(a,b),則方程2x-y+t=0,解得t=8.故所求直線方程為2x-y+8=0.12345678910111213141516177.直線l與l1關(guān)于點(diǎn)(1,-1)中心對(duì)稱,若直線l的方程是2x+3y-6=0,則直線l1的方程是

.

2x+3y+8=0解析

在直線l1上任取一點(diǎn)A(x,y),則點(diǎn)A關(guān)于點(diǎn)(1,-1)的對(duì)稱點(diǎn)B(2-x,-2-y)一定在直線l:2x+3y-6=0上,故有2(2-x)+3(-2-y)-6=0,整理得2x+3y+8=0.故直線l1的方程為2x+3y+8=0.12345678910111213141516178.已知直線l:y=3x+3.求:(1)點(diǎn)P(4,5)關(guān)于直線l的對(duì)稱點(diǎn)坐標(biāo);(2)直線y=x-2關(guān)于直線l對(duì)稱的直線的方程.解

(1)設(shè)點(diǎn)P(4,5)關(guān)于直線l:y=3x+3的對(duì)稱點(diǎn)的坐標(biāo)為(a,b),故點(diǎn)P(4,5)關(guān)于直線l的對(duì)稱點(diǎn)的坐標(biāo)為(-2,7).12345678910111213141516177x+y+22=0,即直線x-y-2=0關(guān)于直線l對(duì)稱的直線的方程為7x+y+22=0.1234567891011121314151617B級(jí)關(guān)鍵能力提升練9.已知點(diǎn)A(5,7)與點(diǎn)B關(guān)于直線l:y=x+1對(duì)稱,則點(diǎn)B的坐標(biāo)為(

)A.(7,6) B.(4,7)C.(6,-7) D.(6,6)D123456789101112131415161710.一束光線從點(diǎn)M(5,3)射出,經(jīng)x軸反射后的光線經(jīng)過(guò)點(diǎn)N(7,3),則反射光線所在的直線方程為(

)A.y=3x-18 B.y=-3x-12C.y=3x+12 D.y=-3x+18A解析

根據(jù)光線的反射原理,入射光線上的點(diǎn)關(guān)于反射面對(duì)稱的點(diǎn)在反射光線的反向延長(zhǎng)線上.由題可知,點(diǎn)M(5,3)關(guān)于x軸對(duì)稱的點(diǎn)為M'(5,-3),則M'(5,-3)在反射光線的反向延長(zhǎng)線上.則kM'N==3,所以反射光線所在的直線方程為y-3=3(x-7),即y=3x-18.故選A.123456789101112131415161711.（多選題）光線從A(-3,4)點(diǎn)射出,到x軸上的B點(diǎn)后,被x軸反射到y(tǒng)軸上的C點(diǎn),又被y軸反射,這時(shí)反射線恰好過(guò)點(diǎn)D(-1,6),則BC所在直線的方程是(

)A.5x-2y+7=0B.3x+y-1=0C.3x-2y+4=0D.2x-y-3=0AC1234567891011121314151617A123456789101112131415161713.已知點(diǎn)A(1,3)關(guān)于直線y=kx+b對(duì)稱的點(diǎn)是B(-2,1),則直線y=kx+b的方程是

.

123456789101112131415161714.將一張畫(huà)了直角坐標(biāo)系(兩坐標(biāo)軸單位長(zhǎng)度相同)的紙折疊一次,使點(diǎn)A(2,0)與點(diǎn)B(-2,4)重合,則與點(diǎn)C(5,8)重合的點(diǎn)的坐標(biāo)是

.

(6,7)解析

由已知得折線為線段AB的垂直平分線,設(shè)垂直平分線的方程為y=kx+b,線段AB的中點(diǎn)為(0,2),斜率為kAB==-1,則線段AB的垂直平分線的斜率k為1,將點(diǎn)(0,2)代入,可得b=2,故垂直平分線的方程為y=x+2.設(shè)點(diǎn)C(5,8)關(guān)于直線y=x+2的對(duì)稱點(diǎn)為P(x0,y0),123456789101112131415161715.已知點(diǎn)A(0,4)與點(diǎn)B關(guān)于直線l0:x+2y-3=