人教A版高中數(shù)學(xué)(選擇性必修第一冊(cè))同步講義第19講 2.3直線的交點(diǎn)坐標(biāo)與距離公式（含解析）

第06講2.3直線的交點(diǎn)坐標(biāo)與距離公式（2.3.1兩條直線的交點(diǎn)坐標(biāo)+2.3.2兩點(diǎn)間的距離公式+2.3.3點(diǎn)到直線的距離公式+2.3.4兩條平行線間的距離公式）課程標(biāo)準(zhǔn)學(xué)習(xí)目標(biāo)①掌握兩條直線的位置關(guān)系中的相交幾何意義，并能根據(jù)已知條件求出兩條直線的交點(diǎn)坐標(biāo)，并能根據(jù)兩條直線相交的性質(zhì)求待定參數(shù)。②會(huì)求平面內(nèi)點(diǎn)與直線的距離，并能解決與距離有關(guān)的平面幾何問(wèn)題。③.會(huì)用兩點(diǎn)間的距離公式求平面內(nèi)兩點(diǎn)間的距離.。④能應(yīng)用公式求兩平行線間的距離，以此解決與平面距離有關(guān)的綜合問(wèn)題。1.會(huì)求兩條直線的交點(diǎn)坐標(biāo)，通過(guò)兩條直線相交的性質(zhì)，解決與直線相交有關(guān)的問(wèn)題；2.掌握利用向量法推導(dǎo)兩點(diǎn)間距離公式的方法，并能用兩點(diǎn)間距離公式求兩點(diǎn)間的距離，以及解決與平面距離相關(guān)的問(wèn)題；3.會(huì)用公式解決與點(diǎn)到直線距離有關(guān)的問(wèn)題，并能解決與之相關(guān)的綜合問(wèn)題；4.熟練應(yīng)用公式求平面內(nèi)兩平行線間的距離，以及與距離有關(guān)的參數(shù)的求解，能處理平面內(nèi)與距離有關(guān)的問(wèn)題.；知識(shí)點(diǎn)01：兩條直線的交點(diǎn)坐標(biāo)直線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）和SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）的公共點(diǎn)的坐標(biāo)與方程組SKIPIF1<0的解一一對(duì)應(yīng).SKIPIF1<0與SKIPIF1<0相交SKIPIF1<0方程組有唯一解，交點(diǎn)坐標(biāo)就是方程組的解；SKIPIF1<0與SKIPIF1<0平行SKIPIF1<0方程組無(wú)解；SKIPIF1<0與SKIPIF1<0重合SKIPIF1<0方程組有無(wú)數(shù)個(gè)解.【即學(xué)即練1】（2023·江蘇·高二假期作業(yè)）分別判斷下列直線SKIPIF1<0與SKIPIF1<0是否相交．如果相交，求出交點(diǎn)的坐標(biāo)．(1)SKIPIF1<0，SKIPIF1<0；(2)SKIPIF1<0，SKIPIF1<0；(3)SKIPIF1<0，SKIPIF1<0．【答案】(1)相交，交點(diǎn)坐標(biāo)為SKIPIF1<0(2)不相交(3)不相交【詳解】（1）解方程組SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0相交，交點(diǎn)坐標(biāo)為SKIPIF1<0．（2）解方程組SKIPIF1<0，方程組無(wú)解，所以SKIPIF1<0與SKIPIF1<0無(wú)公共點(diǎn)，即SKIPIF1<0與SKIPIF1<0不相交．（3）解方程組SKIPIF1<0，因?yàn)榉匠蘏KIPIF1<0可化為SKIPIF1<0，所以方程組有無(wú)數(shù)組解，所以SKIPIF1<0與SKIPIF1<0有無(wú)數(shù)個(gè)公共點(diǎn)，即SKIPIF1<0與SKIPIF1<0不相交．知識(shí)點(diǎn)02：兩點(diǎn)間的距離平面上任意兩點(diǎn)SKIPIF1<0，SKIPIF1<0間的距離公式為SKIPIF1<0特別地，原點(diǎn)SKIPIF1<0與任一點(diǎn)SKIPIF1<0的距離SKIPIF1<0.【即學(xué)即練2】（2023·江蘇·高二假期作業(yè)）已知點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0間的距離為SKIPIF1<0，則SKIPIF1<0________．【答案】9或SKIPIF1<0【詳解】由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．故答案為：9或SKIPIF1<0.知識(shí)點(diǎn)03：點(diǎn)到直線的距離平面上任意一點(diǎn)SKIPIF1<0到直線SKIPIF1<0：SKIPIF1<0的距離SKIPIF1<0.【即學(xué)即練3】（2023春·上海青浦·高二統(tǒng)考期末）點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為_(kāi)_________．【答案】SKIPIF1<0【詳解】由點(diǎn)到直線的距離公式，可得點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0.故答案為：SKIPIF1<0.知識(shí)點(diǎn)04：兩條平行線間的距離一般地，兩條平行直線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）和SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）間的距離SKIPIF1<0.【即學(xué)即練4】（2023秋·廣西河池·高二統(tǒng)考期末）已知直線SKIPIF1<0，SKIPIF1<0相互平行，則SKIPIF1<0、SKIPIF1<0之間的距離為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】因?yàn)橹本€SKIPIF1<0，SKIPIF1<0相互平行，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0、SKIPIF1<0之間的距離SKIPIF1<0.故選：A.知識(shí)點(diǎn)05：對(duì)稱問(wèn)題1、點(diǎn)關(guān)于點(diǎn)對(duì)稱問(wèn)題(方法：中點(diǎn)坐標(biāo)公式)求點(diǎn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0由：SKIPIF1<0SKIPIF1<0SKIPIF1<02、點(diǎn)關(guān)于直線對(duì)稱問(wèn)題（聯(lián)立兩個(gè)方程）求點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0：SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0①設(shè)SKIPIF1<0中點(diǎn)為SKIPIF1<0利用中點(diǎn)坐標(biāo)公式得SKIPIF1<0，將SKIPIF1<0代入直線SKIPIF1<0：SKIPIF1<0中；②SKIPIF1<0整理得：SKIPIF1<0【即學(xué)即練5】（2023秋·高二課時(shí)練習(xí)）若點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱，則SKIPIF1<0_________；SKIPIF1<0__________．【答案】42【詳解】依題意，直線SKIPIF1<0的斜率為SKIPIF1<0，線段SKIPIF1<0的中點(diǎn)SKIPIF1<0，于是SKIPIF1<0，整理得SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0.故答案為：4；23、直線關(guān)于點(diǎn)對(duì)稱問(wèn)題(求SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱直線SKIPIF1<0，則SKIPIF1<0）方法一：在直線SKIPIF1<0上找一點(diǎn)SKIPIF1<0，求點(diǎn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱的點(diǎn)SKIPIF1<0，根據(jù)SKIPIF1<0，再由點(diǎn)斜式求解；方法二：由SKIPIF1<0SKIPIF1<0，設(shè)出SKIPIF1<0的直線方程，由點(diǎn)SKIPIF1<0到兩直線的距離相等SKIPIF1<0求參數(shù).方法三：在直線SKIPIF1<0任意一點(diǎn)SKIPIF1<0，求該點(diǎn)關(guān)于點(diǎn)SKIPIF1<0對(duì)稱的點(diǎn)SKIPIF1<0，則該點(diǎn)SKIPIF1<0在直線SKIPIF1<0上.【即學(xué)即練6】（2023·高二單元測(cè)試）直線SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱直線方程是______．【答案】SKIPIF1<0【詳解】設(shè)對(duì)稱直線為SKIPIF1<0，則有SKIPIF1<0，即SKIPIF1<0解這個(gè)方程得SKIPIF1<0（舍）或SKIPIF1<0.所以對(duì)稱直線SKIPIF1<0的方程中SKIPIF1<0.故答案為：SKIPIF1<0.4、直線關(guān)于直線對(duì)稱問(wèn)題4.1直線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）和SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）相交,求SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱直線SKIPIF1<0①求出SKIPIF1<0與SKIPIF1<0的交點(diǎn)SKIPIF1<0②在SKIPIF1<0上任意取一點(diǎn)SKIPIF1<0(非SKIPIF1<0點(diǎn)），求出SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0③根據(jù)SKIPIF1<0，SKIPIF1<0兩點(diǎn)求出直線SKIPIF1<04.2直線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）和SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）平行,求SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱直線SKIPIF1<0①SKIPIF1<0②在直線SKIPIF1<0上任取一點(diǎn)SKIPIF1<0，求點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0，利用點(diǎn)斜式求直線SKIPIF1<0.【即學(xué)即練7】（2023·高二課時(shí)練習(xí)）求直線SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱的直線SKIPIF1<0的方程．【答案】SKIPIF1<0【詳解】聯(lián)立兩直線方程SKIPIF1<0，解得SKIPIF1<0，即兩直線的交點(diǎn)為SKIPIF1<0，取直線SKIPIF1<0：SKIPIF1<0上一點(diǎn)SKIPIF1<0，設(shè)其關(guān)于直線SKIPIF1<0：SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，因?yàn)樗笾本€過(guò)SKIPIF1<0，SKIPIF1<0，方程為SKIPIF1<0，即SKIPIF1<0.【即學(xué)即練8】（2023春·上海寶山·高二上海市吳淞中學(xué)校考期中）直線SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱的直線方程為_(kāi)_______【答案】SKIPIF1<0【詳解】設(shè)所求直線方程為SKIPIF1<0，且SKIPIF1<0，直線SKIPIF1<0與直線SKIPIF1<0間的距離為SKIPIF1<0，則直線SKIPIF1<0與直線SKIPIF1<0間的距離為SKIPIF1<0，又SKIPIF1<0，得SKIPIF1<0，所以所求直線方程為SKIPIF1<0，故答案為：SKIPIF1<0.題型01求直線交點(diǎn)坐標(biāo)【典例1】（2023·江蘇·高二假期作業(yè)）直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)坐標(biāo)是（

）A．(2,0) B．(2,1)C．(0,2) D．(1,2)【答案】C【詳解】解方程組SKIPIF1<0得SKIPIF1<0，即直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)坐標(biāo)是(0,2)．故選：C.【典例2】（2023秋·高二課時(shí)練習(xí)）若直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)位于第一象限，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】聯(lián)立SKIPIF1<0得SKIPIF1<0，因?yàn)橹本€SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)位于第一象限，所以SKIPIF1<0，解得SKIPIF1<0.故選：D【變式1】（2023秋·天津·高二校聯(lián)考期末）過(guò)直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)，且與直線SKIPIF1<0垂直的直線方程是（

）．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】聯(lián)立方程SKIPIF1<0，解得SKIPIF1<0，所以交點(diǎn)坐標(biāo)為SKIPIF1<0；直線SKIPIF1<0的斜率為SKIPIF1<0，所以所求直線方程的斜率為SKIPIF1<0，由點(diǎn)斜式直線方程得：所求直線方程為SKIPIF1<0，即SKIPIF1<0；故選：B.【變式2】（2023·高二課時(shí)練習(xí)）若直線SKIPIF1<0與直線SKIPIF1<0相交且交點(diǎn)在第二象限內(nèi)，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】若直線SKIPIF1<0與直線SKIPIF1<0平行或重合，則SKIPIF1<0，解得SKIPIF1<0，若直線SKIPIF1<0與直線SKIPIF1<0相交，可得SKIPIF1<0且SKIPIF1<0，則有：聯(lián)立方程SKIPIF1<0，解得SKIPIF1<0，即交點(diǎn)坐標(biāo)SKIPIF1<0，由題意可得：SKIPIF1<0，解得SKIPIF1<0；綜上所述：k的取值范圍為SKIPIF1<0.故選：C.題型02由方程組解的個(gè)數(shù)判斷直線的位置關(guān)系【典例1】（2023秋·高二課時(shí)練習(xí)）判斷下列各對(duì)直線的位置關(guān)系．如果相交，求出交點(diǎn)坐標(biāo)．(1)直線SKIPIF1<0；(2)直線SKIPIF1<0．【答案】(1)相交，交點(diǎn)是SKIPIF1<0(2)答案見(jiàn)解析【詳解】（1）聯(lián)立SKIPIF1<0，解得SKIPIF1<0，所以兩直線相交，交點(diǎn)坐標(biāo)為SKIPIF1<0.（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，聯(lián)立SKIPIF1<0，方程組有無(wú)數(shù)組解，故兩直線重合，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，聯(lián)立SKIPIF1<0，方程組無(wú)解，故兩直線平行，當(dāng)SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，所以兩直線相交，交點(diǎn)坐標(biāo)為SKIPIF1<0.綜上所述：當(dāng)SKIPIF1<0時(shí)，兩直線重合；當(dāng)SKIPIF1<0時(shí)，兩直線平行；當(dāng)SKIPIF1<0時(shí)，兩直線相交，交點(diǎn)坐標(biāo)為SKIPIF1<0.【典例2】（2022·上?！じ呷龑ｎ}練習(xí)）若關(guān)于SKIPIF1<0、SKIPIF1<0的方程組SKIPIF1<0無(wú)解，則實(shí)數(shù)SKIPIF1<0________【答案】SKIPIF1<0【詳解】由題意關(guān)于SKIPIF1<0、SKIPIF1<0的方程組SKIPIF1<0無(wú)解，即直線SKIPIF1<0和直線SKIPIF1<0平行，故SKIPIF1<0，所以SKIPIF1<0，此時(shí)直線SKIPIF1<0即SKIPIF1<0，確實(shí)與SKIPIF1<0平行，故滿足題意，所以實(shí)數(shù)SKIPIF1<0.故答案為：-2.【變式1】（2022·高二課時(shí)練習(xí)）若關(guān)于SKIPIF1<0的二元一次方程組SKIPIF1<0有無(wú)窮多組解，則SKIPIF1<0______．【答案】SKIPIF1<0【詳解】依題意二元一次方程組SKIPIF1<0有無(wú)窮多組解，即兩個(gè)方程對(duì)應(yīng)的直線重合，由SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，二元一次方程組為SKIPIF1<0，兩直線不重合，不符合題意.當(dāng)SKIPIF1<0時(shí)，二元一次方程組為SKIPIF1<0，兩直線重合，符合題意.綜上所述，SKIPIF1<0的值為SKIPIF1<0.故答案為：SKIPIF1<0【變式2】（2022·高二課時(shí)練習(xí)）關(guān)于SKIPIF1<0?SKIPIF1<0的二元一次方程組SKIPIF1<0有無(wú)窮多組解，則SKIPIF1<0與SKIPIF1<0的積是_____.【答案】-35【詳解】因?yàn)閤?y的二元一次方程組SKIPIF1<0有無(wú)窮多組解，所以直線SKIPIF1<0與直線SKIPIF1<0重合，所以SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，故答案為：-35題型03由直線交點(diǎn)的個(gè)數(shù)求參數(shù)【典例1】（2022秋·廣東廣州·高二廣州市第一一三中學(xué)?？茧A段練習(xí)）直線SKIPIF1<0與直線SKIPIF1<0相交，則實(shí)數(shù)SKIPIF1<0的值為（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0 C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0且SKIPIF1<0【答案】D【詳解】因直線SKIPIF1<0與直線SKIPIF1<0相交，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0且SKIPIF1<0，所以實(shí)數(shù)k的值為SKIPIF1<0且SKIPIF1<0.故選：D【典例2】（2022·高二校聯(lián)考課時(shí)練習(xí)）若關(guān)于SKIPIF1<0，SKIPIF1<0的方程組SKIPIF1<0有唯一解，則實(shí)數(shù)SKIPIF1<0滿足的條件是________.【答案】SKIPIF1<0/SKIPIF1<0【詳解】由SKIPIF1<0，可得SKIPIF1<0，由關(guān)于SKIPIF1<0，SKIPIF1<0的方程組SKIPIF1<0有唯一解，可得方程SKIPIF1<0有唯一解，則SKIPIF1<0故答案為：SKIPIF1<0【典例3】（2022·高二校聯(lián)考課時(shí)練習(xí)）已知三條直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.（1）若直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0交于一點(diǎn)，求實(shí)數(shù)SKIPIF1<0的值；（2）若直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0不能圍成三角形，求實(shí)數(shù)SKIPIF1<0的值.【答案】（1）SKIPIF1<0或SKIPIF1<0；（2）SKIPIF1<0或SKIPIF1<0或4或SKIPIF1<0.【詳解】（1）∵直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0交于一點(diǎn)，∴SKIPIF1<0與SKIPIF1<0不平行，∴SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0的交點(diǎn)為SKIPIF1<0，代入SKIPIF1<0的方程，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.（2）若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0交于一點(diǎn)，則SKIPIF1<0或SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則SKIPIF1<0；若SKIPIF1<0，則不存在滿足條件的實(shí)數(shù)SKIPIF1<0.綜上，可得SKIPIF1<0或SKIPIF1<0或4或SKIPIF1<0.【變式1】（2022·江蘇·高二專題練習(xí)）若三條直線SKIPIF1<0，SKIPIF1<0與SKIPIF1<0共有兩個(gè)交點(diǎn)，則實(shí)數(shù)SKIPIF1<0的值為（

）A．1 B．-2 C．1或-2 D．-1【答案】C【詳解】由題意可得三條直線中，有兩條直線互相平行，∵直線SKIPIF1<0和直線SKIPIF1<0不平行，∴直線SKIPIF1<0和直線SKIPIF1<0平行或直線SKIPIF1<0和直線SKIPIF1<0平行，∵直線SKIPIF1<0的斜率為1，直線SKIPIF1<0的斜率為SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0.故選：C.【變式2】（2022·高二課時(shí)練習(xí)）三條直線SKIPIF1<0?SKIPIF1<0?SKIPIF1<0有且只有兩個(gè)交點(diǎn)，求實(shí)數(shù)SKIPIF1<0的值.【答案】SKIPIF1<0或SKIPIF1<0【詳解】由SKIPIF1<0得：SKIPIF1<0，即SKIPIF1<0有一個(gè)交點(diǎn)SKIPIF1<0，SKIPIF1<0或SKIPIF1<0；即SKIPIF1<0或SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0.題型04由直線的交點(diǎn)坐標(biāo)求參數(shù)【典例1】（2023秋·高一單元測(cè)試）若直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)在第四象限，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由方程組SKIPIF1<0，解得SKIPIF1<0，即兩直線的交點(diǎn)坐標(biāo)為SKIPIF1<0，因?yàn)閮芍本€的交點(diǎn)位于第四象限，可得SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故選：D.【典例2】（2023·高二課時(shí)練習(xí)）若直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)在第一象限，則實(shí)數(shù)SKIPIF1<0的取值范圍是___________.【答案】SKIPIF1<0【詳解】由題意，直線SKIPIF1<0，令SKIPIF1<0，可得SKIPIF1<0；令SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0，如圖所示，當(dāng)直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，可得SKIPIF1<0；當(dāng)直線SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，可得SKIPIF1<0，要使得直線SKIPIF1<0與直線SKIPIF1<0的交點(diǎn)在第一象限，則SKIPIF1<0，即實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.【變式1】（2023·江蘇·高二假期作業(yè)）若三條直線SKIPIF1<0和SKIPIF1<0交于一點(diǎn)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．SKIPIF1<0【答案】C【詳解】解：聯(lián)立SKIPIF1<0得SKIPIF1<0.把SKIPIF1<0代入SKIPIF1<0得SKIPIF1<0.故選：C【變式2】（2023·江蘇·高二假期作業(yè)）兩直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)在SKIPIF1<0軸上，則SKIPIF1<0的值是（

）A．-24 B．6 C．±6 D．24【答案】C【詳解】因?yàn)閮蓷l直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)在SKIPIF1<0軸上，所以設(shè)交點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，消去SKIPIF1<0，可得SKIPIF1<0．故選：SKIPIF1<0．題型05三線圍成三角形問(wèn)題【典例1】（2023秋·高二課時(shí)練習(xí)）使三條直線SKIPIF1<0不能圍成三角形的實(shí)數(shù)SKIPIF1<0的值最多有幾個(gè)（

）A．3個(gè) B．4個(gè) C．5個(gè) D．6個(gè)【答案】B【詳解】要使三條直線不能圍成三角形，存在兩條直線平行或三條直線交于一點(diǎn)，若SKIPIF1<0平行，則SKIPIF1<0，即SKIPIF1<0；若SKIPIF1<0平行，則SKIPIF1<0，即無(wú)解；若SKIPIF1<0平行，則SKIPIF1<0，即SKIPIF1<0；若三條直線交于一點(diǎn)，SKIPIF1<0，可得SKIPIF1<0或SKIPIF1<0；經(jīng)檢驗(yàn)知：SKIPIF1<0均滿足三條直線不能圍成三角形，故m最多有4個(gè).故選：B【典例2】（2023·江蘇·高二假期作業(yè)）若三條直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0能構(gòu)成三角形，求SKIPIF1<0應(yīng)滿足的條件．

【答案】SKIPIF1<0且SKIPIF1<0【詳解】為使三條直線能構(gòu)成三角形，需三條直線兩兩相交且不共點(diǎn)．①若SKIPIF1<0，則由SKIPIF1<0，得SKIPIF1<0；②若SKIPIF1<0，則由SKIPIF1<0，得SKIPIF1<0；③若SKIPIF1<0，則由SKIPIF1<0，得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0與SKIPIF1<0三線重合，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0平行．④若三條直線交于一點(diǎn)，由SKIPIF1<0，解得SKIPIF1<0，將SKIPIF1<0的交點(diǎn)SKIPIF1<0的坐標(biāo)代入SKIPIF1<0的方程，解得SKIPIF1<0(舍去)，或SKIPIF1<0，所以要使三條直線能構(gòu)成三角形，需SKIPIF1<0且SKIPIF1<0．【變式1】（多選）（2023·全國(guó)·高二專題練習(xí)）三條直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0構(gòu)成三角形，則SKIPIF1<0的值不能為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．－2【答案】AC【詳解】直線SKIPIF1<0與SKIPIF1<0都經(jīng)過(guò)原點(diǎn)，而無(wú)論SKIPIF1<0為何值，直線SKIPIF1<0總不經(jīng)過(guò)原點(diǎn)，因此，要滿足三條直線構(gòu)成三角形，只需直線SKIPIF1<0與另兩條直線不平行，所以SKIPIF1<0．故選：AC.【變式2】（2023秋·浙江寧波·高二期末）若三條直線SKIPIF1<0與SKIPIF1<0能圍成一個(gè)直角三角形，則SKIPIF1<0__________．【答案】SKIPIF1<0或1【詳解】顯然，3x-y+1=0，x+y+3=0有交點(diǎn)，若SKIPIF1<0與SKIPIF1<0垂直，則SKIPIF1<0；若SKIPIF1<0與SKIPIF1<0垂直，則SKIPIF1<0．所以SKIPIF1<0或1．故答案為：SKIPIF1<0或1題型06直線交點(diǎn)系方程及其應(yīng)用【典例1】（2023·江蘇·高二假期作業(yè)）設(shè)直線SKIPIF1<0經(jīng)過(guò)SKIPIF1<0和SKIPIF1<0的交點(diǎn)，且與兩坐標(biāo)軸圍成等腰直角三角形，則直線SKIPIF1<0的方程為_(kāi)__________.【答案】SKIPIF1<0或SKIPIF1<0【詳解】方法一：由SKIPIF1<0，得SKIPIF1<0，所以兩條直線的交點(diǎn)坐標(biāo)為（14，10）,由題意可得直線SKIPIF1<0的斜率為1或-1，所以直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.方法二：設(shè)直線SKIPIF1<0的方程為SKIPIF1<0，整理得SKIPIF1<0，由題意，得SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0.【典例2】（2022·高二課時(shí)練習(xí)）已知兩直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)為SKIPIF1<0．求：(1)過(guò)點(diǎn)SKIPIF1<0與SKIPIF1<0的直線方程；(2)過(guò)點(diǎn)SKIPIF1<0且與直線SKIPIF1<0平行的直線方程．【答案】(1)SKIPIF1<0(2)SKIPIF1<0（1）設(shè)過(guò)直線SKIPIF1<0和SKIPIF1<0交點(diǎn)的直線方程為SKIPIF1<0，即SKIPIF1<0．①把點(diǎn)SKIPIF1<0代入方程①，化簡(jiǎn)得SKIPIF1<0，解得SKIPIF1<0，所以過(guò)點(diǎn)P與Q的直線方程為SKIPIF1<0，即SKIPIF1<0．（2）由兩直線平行，得SKIPIF1<0，得SKIPIF1<0，所以所求直線的方程為SKIPIF1<0，即SKIPIF1<0．【變式1】（2022秋·高二課時(shí)練習(xí)）過(guò)兩直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)和原點(diǎn)的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】設(shè)過(guò)兩直線交點(diǎn)的直線系方程為SKIPIF1<0，代入原點(diǎn)坐標(biāo)，得SKIPIF1<0，解得SKIPIF1<0，故所求直線方程為SKIPIF1<0，即SKIPIF1<0.故選：D.【變式2】（2022·高二單元測(cè)試）已知直線SKIPIF1<0：SKIPIF1<0（SKIPIF1<0）．求證：直線SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0，并求點(diǎn)SKIPIF1<0的坐標(biāo)．【答案】證明見(jiàn)解析，SKIPIF1<0SKIPIF1<0【詳解】證明：原方程整理為SKIPIF1<0，則由SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0點(diǎn)坐標(biāo)為SKIPIF1<0．【變式3】（2022·高二課時(shí)練習(xí)）直線SKIPIF1<0經(jīng)過(guò)直線SKIPIF1<0的交點(diǎn)，且與坐標(biāo)軸圍成的三角形是等腰直角三角形，求直線SKIPIF1<0的方程.【答案】SKIPIF1<0或SKIPIF1<0【詳解】解：設(shè)直線SKIPIF1<0方程為SKIPIF1<0SKIPIF1<0，化簡(jiǎn)得SKIPIF1<0,SKIPIF1<0直線SKIPIF1<0與坐標(biāo)軸圍成的三角形是等腰直角三角形，SKIPIF1<0直線SKIPIF1<0的斜率為SKIPIF1<0,SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.代入并化簡(jiǎn)得直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0.所以所求的直線方程為SKIPIF1<0或SKIPIF1<0.題型07求兩點(diǎn)間的距離公式【典例1】（2023·江蘇·高二假期作業(yè)）已知SKIPIF1<0，SKIPIF1<0兩點(diǎn)分別在兩條互相垂直的直線SKIPIF1<0和SKIPIF1<0上，且SKIPIF1<0線段的中點(diǎn)為SKIPIF1<0，則線段SKIPIF1<0的長(zhǎng)為（

）A．11 B．10 C．9 D．8【答案】B【詳解】因?yàn)橹本€SKIPIF1<0和SKIPIF1<0互相垂直，所以SKIPIF1<0，解得SKIPIF1<0，所以線段AB的中點(diǎn)為SKIPIF1<0，所以設(shè)SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，故選：C【典例2】（2023·全國(guó)·高三專題練習(xí)）已知直線SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，直線SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，SKIPIF1<0與SKIPIF1<0相交于點(diǎn)SKIPIF1<0，則SKIPIF1<0（

）A．10 B．13 C．16 D．20【答案】B【詳解】解：因?yàn)镾KIPIF1<0，所以直線SKIPIF1<0與直線SKIPIF1<0互相垂直且垂足為點(diǎn)SKIPIF1<0，又因?yàn)橹本€SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，直線SKIPIF1<0，即SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，所以在SKIPIF1<0中，SKIPIF1<0，故選：B.【變式1】（2023秋·高二課時(shí)練習(xí)）已知SKIPIF1<0，點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，且SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】因?yàn)辄c(diǎn)C在x軸上，設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，化簡(jiǎn)可得：SKIPIF1<0，所以SKIPIF1<0.故選：D.【變式2】（2023·江蘇·高二假期作業(yè)）直線SKIPIF1<0和直線SKIPIF1<0分別過(guò)定點(diǎn)SKIPIF1<0和SKIPIF1<0，則SKIPIF1<0|________．【答案】SKIPIF1<0【詳解】將直線SKIPIF1<0的方程變形為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，即點(diǎn)SKIPIF1<0，將直線SKIPIF1<0的方程變形為SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，即點(diǎn)SKIPIF1<0，所以，SKIPIF1<0.故答案為：SKIPIF1<0.【變式3】（2023·高三課時(shí)練習(xí)）如圖，SKIPIF1<0是邊長(zhǎng)為1的正三角形，SKIPIF1<0，SKIPIF1<0分別為線段SKIPIF1<0，SKIPIF1<0上一點(diǎn)，滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0與SKIPIF1<0的交點(diǎn)為SKIPIF1<0，則線段SKIPIF1<0的長(zhǎng)度為_(kāi)__________.【答案】SKIPIF1<0【詳解】解：以SKIPIF1<0為原點(diǎn)，SKIPIF1<0為SKIPIF1<0軸，建立如圖所示的平面直角坐標(biāo)系，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0，聯(lián)立SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<0.題型08距離公式的應(yīng)用【典例1】（2023春·江西·高三校聯(lián)考開(kāi)學(xué)考試）費(fèi)馬點(diǎn)是指三角形內(nèi)到三角形三個(gè)頂點(diǎn)距離之和最小的點(diǎn).當(dāng)三角形三個(gè)內(nèi)角均小于120°時(shí)，費(fèi)馬點(diǎn)與三個(gè)頂點(diǎn)連線正好三等分費(fèi)馬點(diǎn)所在的周角，即該點(diǎn)所對(duì)的三角形三邊的張角相等且均為120°.根據(jù)以上性質(zhì)，.則SKIPIF1<0的最小值為（

）A．4 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】由題意得：SKIPIF1<0的幾何意義為點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離之和的最小值，因?yàn)镾KIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故三角形ABC為等腰直角三角形，，取SKIPIF1<0的中點(diǎn)SKIPIF1<0，連接SKIPIF1<0，與SKIPIF1<0交于點(diǎn)SKIPIF1<0，連接SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故SKIPIF1<0，則SKIPIF1<0，故點(diǎn)SKIPIF1<0到三角形三個(gè)頂點(diǎn)距離之和最小，即SKIPIF1<0取得最小值，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，同理得：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0的最小值為SKIPIF1<0.故選：B【典例2】（2022秋·福建·高二校聯(lián)考期中）著名數(shù)學(xué)家華羅庚曾說(shuō)過(guò)：“數(shù)形結(jié)合百般好，割裂分家萬(wàn)事休．”事實(shí)上，有很多代數(shù)問(wèn)題可以轉(zhuǎn)化為幾何問(wèn)題加以解決，如：SKIPIF1<0可以轉(zhuǎn)化為點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離，則SKIPIF1<0的最小值為（

）．A．3 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】SKIPIF1<0,可以看作點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離之和，作點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)SKIPIF1<0，顯然當(dāng)SKIPIF1<0三點(diǎn)共線時(shí)，取到最小值，最小值為SKIPIF1<0間的距離SKIPIF1<0.故選：D.【典例3】（2022秋·甘肅嘉峪關(guān)·高二?？计谥校┖瘮?shù)SKIPIF1<0的最小值是_____________.【答案】5【詳解】解：因?yàn)镾KIPIF1<0SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0表示點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0，SKIPIF1<0兩點(diǎn)的距離之和，即SKIPIF1<0，點(diǎn)SKIPIF1<0是SKIPIF1<0軸上的點(diǎn)，則點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)為SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0的最小值是SKIPIF1<0.故答案為：SKIPIF1<0【變式1】（2023·全國(guó)·高三專題練習(xí)）費(fèi)馬點(diǎn)是指三角形內(nèi)到三角形三個(gè)頂點(diǎn)距離之和最小的點(diǎn).當(dāng)三角形三個(gè)內(nèi)角均小于SKIPIF1<0時(shí)，費(fèi)馬點(diǎn)與三個(gè)頂點(diǎn)連線正好三等分費(fèi)馬點(diǎn)所在的周角，即該點(diǎn)所對(duì)的三角形三邊的張角相等均為SKIPIF1<0.根據(jù)以上性質(zhì)，SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【詳解】由題SKIPIF1<0的幾何意義為點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離之和的最小值.由題可知,此時(shí)SKIPIF1<0,且SKIPIF1<0在SKIPIF1<0軸上.故SKIPIF1<0.SKIPIF1<0,SKIPIF1<0.故SKIPIF1<0的最小值為SKIPIF1<0故選：D【變式2】（2022秋·北京·高二北京工業(yè)大學(xué)附屬中學(xué)?？计谥校┲麛?shù)學(xué)家華羅庚曾說(shuō)過(guò)：“數(shù)無(wú)形時(shí)少直覺(jué)，形少數(shù)時(shí)難入微.”事實(shí)上，有很多代數(shù)問(wèn)題可以轉(zhuǎn)化為幾何問(wèn)題加以解決，如：SKIPIF1<0可以轉(zhuǎn)化為平面上點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0的距離.結(jié)合上述觀點(diǎn)，可得SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【詳解】因?yàn)镾KIPIF1<0，記點(diǎn)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)點(diǎn)SKIPIF1<0為線段SKIPIF1<0與SKIPIF1<0軸的交點(diǎn)時(shí)，等號(hào)成立，即SKIPIF1<0的最小值為SKIPIF1<0.故選：C.【變式3】（2023·江蘇·高二假期作業(yè)）某同學(xué)在研究函數(shù)SKIPIF1<0的性質(zhì)時(shí)，聯(lián)想到兩點(diǎn)間的距離公式，從而將函數(shù)變形為SKIPIF1<0，求得SKIPIF1<0的最小值為_(kāi)_______．【答案】SKIPIF1<0【詳解】由變形所得函數(shù)知：SKIPIF1<0表示x軸上的動(dòng)點(diǎn)SKIPIF1<0到兩定點(diǎn)SKIPIF1<0的距離之和，∴當(dāng)且僅當(dāng)SKIPIF1<0與SKIPIF1<0重合時(shí)，SKIPIF1<0有最小值為SKIPIF1<0.故答案為：SKIPIF1<0題型09求點(diǎn)到直線的距離【典例1】（2023·重慶·高二統(tǒng)考學(xué)業(yè)考試）點(diǎn)(1,1)到直線SKIPIF1<0的距離是（

）A．1 B．2 C．SKIPIF1<0【答案】A【詳解】SKIPIF1<0，故選：A【典例2】（2023春·上海浦東新·高二統(tǒng)考期中）已知?jiǎng)狱c(diǎn)SKIPIF1<0在直線SKIPIF1<0上，則SKIPIF1<0的最小值為_(kāi)________.【答案】2【詳解】因?yàn)镾KIP