第37練直線(xiàn)的傾斜角與斜率、直線(xiàn)的方程（精練）【A組

在基礎(chǔ)中考查功底】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0的傾斜角為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)傾斜角和斜率的關(guān)系求解.【詳解】由已知得SKIPIF1<0，故直線(xiàn)斜率SKIPIF1<0由于傾斜的范圍是SKIPIF1<0，則傾斜角為SKIPIF1<0.故選：B.2．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0，則直線(xiàn)SKIPIF1<0的傾斜角為（

）A．30° B．60° C．120° D．135°【答案】B【分析】先由SKIPIF1<0，SKIPIF1<0求斜率，再求傾斜角.【詳解】設(shè)直線(xiàn)SKIPIF1<0的斜率為k,則SKIPIF1<0.令直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.故選：B3．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，且SKIPIF1<0在SKIPIF1<0軸上的截距為SKIPIF1<0，則直線(xiàn)SKIPIF1<0的方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】首先求出直線(xiàn)的斜率，再根據(jù)斜截式計(jì)算可得；【詳解】解：因?yàn)橹本€(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，所以直線(xiàn)SKIPIF1<0的斜率SKIPIF1<0，又直線(xiàn)SKIPIF1<0在SKIPIF1<0軸上的截距為SKIPIF1<0，所以直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0；故選：C4．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0（A，B不同時(shí)為SKIPIF1<0），則下列說(shuō)法中錯(cuò)誤的是（

）A．當(dāng)SKIPIF1<0時(shí)，直線(xiàn)l總與x軸相交B．當(dāng)SKIPIF1<0時(shí)，直線(xiàn)l經(jīng)過(guò)坐標(biāo)原點(diǎn)OC．當(dāng)SKIPIF1<0時(shí)，直線(xiàn)l是x軸所在直線(xiàn)D．當(dāng)SKIPIF1<0時(shí)，直線(xiàn)l不可能與兩坐標(biāo)軸同時(shí)相交【答案】D【分析】根據(jù)直線(xiàn)的知識(shí)對(duì)選項(xiàng)進(jìn)行分析，從而確定正確答案.【詳解】依題意，直線(xiàn)SKIPIF1<0（A，B不同時(shí)為SKIPIF1<0）.A選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，直線(xiàn)方程可化為SKIPIF1<0，此時(shí)直線(xiàn)SKIPIF1<0總與SKIPIF1<0軸有交點(diǎn)，A選項(xiàng)正確.B選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，直線(xiàn)方程為SKIPIF1<0，此時(shí)直線(xiàn)SKIPIF1<0經(jīng)過(guò)原點(diǎn)SKIPIF1<0，B選項(xiàng)正確.C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，直線(xiàn)方程可化為SKIPIF1<0，此時(shí)直線(xiàn)l是x軸所在直線(xiàn)，C選項(xiàng)正確.D選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，如SKIPIF1<0，直線(xiàn)SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0，即直線(xiàn)SKIPIF1<0與兩坐標(biāo)軸同時(shí)相交，D選項(xiàng)錯(cuò)誤.故選：D.5．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)l：SKIPIF1<0的傾斜角為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．－1【答案】A【分析】由傾斜角求出斜率，列方程即可求出m.【詳解】因?yàn)橹本€(xiàn)l的傾斜角為SKIPIF1<0，所以斜率SKIPIF1<0.所以SKIPIF1<0，解得：SKIPIF1<0.故選：A6．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若一次函數(shù)SKIPIF1<0所表示直線(xiàn)的傾斜角為SKIPIF1<0，則SKIPIF1<0的值為（

）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由題意可得SKIPIF1<0，化簡(jiǎn)SKIPIF1<0代入計(jì)算．【詳解】SKIPIF1<0的斜率為SKIPIF1<0即SKIPIF1<0SKIPIF1<0故選：D．7．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0，若直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0的夾角是60°，則k的值為（

）A．SKIPIF1<0或0 B．SKIPIF1<0或0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先求出SKIPIF1<0的傾斜角為120°，再求出直線(xiàn)SKIPIF1<0的傾斜角為0°或60°，直接求斜率k.【詳解】直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，所以?xún)A斜角為120°.要使直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0的夾角是60°，只需直線(xiàn)SKIPIF1<0的傾斜角為0°或60°，所以k的值為0或SKIPIF1<0.故選：A8．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)直線(xiàn)l的方程為SKIPIF1<0，則直線(xiàn)l的傾斜角SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】當(dāng)SKIPIF1<0時(shí)，可得傾斜角為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由直線(xiàn)方程可得斜率SKIPIF1<0，然后由余弦函數(shù)和正切函數(shù)的性質(zhì)求解即可【詳解】當(dāng)SKIPIF1<0時(shí)，方程變?yōu)镾KIPIF1<0，其傾斜角為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由直線(xiàn)方程可得斜率SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，由上知，傾斜角的范圍是SKIPIF1<0．故選：C．9．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0，點(diǎn)SKIPIF1<0，SKIPIF1<0，若直線(xiàn)SKIPIF1<0與線(xiàn)段AB有公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】A【分析】若直線(xiàn)SKIPIF1<0與線(xiàn)段SKIPIF1<0有公共點(diǎn)，由SKIPIF1<0、SKIPIF1<0在直線(xiàn)SKIPIF1<0的兩側(cè)（也可以點(diǎn)在直線(xiàn)上），得SKIPIF1<0（SKIPIF1<0）可得結(jié)論．【詳解】若直線(xiàn)SKIPIF1<0與線(xiàn)段SKIPIF1<0有公共點(diǎn)，則SKIPIF1<0、SKIPIF1<0在直線(xiàn)SKIPIF1<0的兩側(cè)（也可以點(diǎn)在直線(xiàn)上）.令SKIPIF1<0，則有SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0.解得SKIPIF1<0，故選：A.10．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)的傾斜角的范圍是SKIPIF1<0，則此直線(xiàn)的斜率k的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用直線(xiàn)斜率的定義結(jié)合正切函數(shù)的性質(zhì)即可計(jì)算作答.【詳解】當(dāng)直線(xiàn)的傾斜角SKIPIF1<0時(shí)，直線(xiàn)的斜率SKIPIF1<0，因SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0，所以直線(xiàn)的斜率k的取值范圍是SKIPIF1<0.故選：D11．（2023·北京西城·北師大實(shí)驗(yàn)中學(xué)校考三模）設(shè)SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0和過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0交于點(diǎn)SKIPIF1<0，則SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【詳解】試題分析：易得SKIPIF1<0.設(shè)SKIPIF1<0，則消去SKIPIF1<0得：SKIPIF1<0，所以點(diǎn)P在以AB為直徑的圓上，SKIPIF1<0，所以SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0.所以SKIPIF1<0，SKIPIF1<0.選B.法二、因?yàn)閮芍本€(xiàn)的斜率互為負(fù)倒數(shù)，所以SKIPIF1<0，點(diǎn)P的軌跡是以AB為直徑的圓.以下同法一.二、多選題12．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0,則下列命題正確的是（

）A．直線(xiàn)的傾斜角是SKIPIF1<0B．無(wú)論SKIPIF1<0如何變化,直線(xiàn)不過(guò)原點(diǎn)C．直線(xiàn)的斜率一定存在D．當(dāng)直線(xiàn)和兩坐標(biāo)軸都相交時(shí),它和坐標(biāo)軸圍成的三角形的面積不小于1【答案】BD【分析】根據(jù)直線(xiàn)方程考慮SKIPIF1<0的值，當(dāng)取SKIPIF1<0時(shí)，顯然選項(xiàng)A錯(cuò)誤；將原點(diǎn)代入直線(xiàn)方程；可知選項(xiàng)B正確，當(dāng)SKIPIF1<0時(shí)選項(xiàng)C錯(cuò)誤；求出直線(xiàn)和兩坐標(biāo)軸的交點(diǎn)，求出面積范圍即可判斷選項(xiàng)D正誤.【詳解】解:由題知,直線(xiàn)SKIPIF1<0，若SKIPIF1<0，則直線(xiàn)為SKIPIF1<0，傾斜角為SKIPIF1<0,與選項(xiàng)A不符,故選項(xiàng)A錯(cuò)誤，將原點(diǎn)SKIPIF1<0代入直線(xiàn)方程可得SKIPIF1<0不符，故選項(xiàng)B正確，若SKIPIF1<0,則直線(xiàn)為SKIPIF1<0,斜率不存在,故選項(xiàng)C錯(cuò)誤，當(dāng)直線(xiàn)和兩坐標(biāo)軸都相交時(shí)，交點(diǎn)為SKIPIF1<0，它和坐標(biāo)軸圍成的三角形的面積為SKIPIF1<0，SKIPIF1<0，故選項(xiàng)D正確，故選:BD13．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，直線(xiàn)l的方程為SKIPIF1<0，則直線(xiàn)l的傾斜角可能為（

）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】CD【分析】對(duì)SKIPIF1<0分類(lèi)討論結(jié)合斜率與傾斜角的關(guān)系即得.【詳解】當(dāng)SKIPIF1<0時(shí)，則直線(xiàn)的斜率為SKIPIF1<0，所以直線(xiàn)的傾斜角可能為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則直線(xiàn)的斜率不存在，所以直線(xiàn)的傾斜角為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，則直線(xiàn)的斜率為SKIPIF1<0，所以直線(xiàn)的傾斜角范圍為SKIPIF1<0，不可能為0和SKIPIF1<0.故選：CD.三、填空題14．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若某直線(xiàn)經(jīng)過(guò)A（SKIPIF1<0，SKIPIF1<0），B（1，SKIPIF1<0SKIPIF1<0）兩點(diǎn)，則此直線(xiàn)的傾斜角為.【答案】120°【分析】利用斜率公式求得斜率，進(jìn)而得到傾斜角.【詳解】直線(xiàn)的斜率SKIPIF1<0,故傾斜角SKIPIF1<0，故答案為：120°.15．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)l的斜率為k，傾斜角為α，而α∈SKIPIF1<0，則k的取值范圍是．【答案】SKIPIF1<0.【解析】由直線(xiàn)傾斜角的范圍再結(jié)合正切函數(shù)的單調(diào)性即可求出k的取值范圍.【詳解】當(dāng)SKIPIF1<0≤α<SKIPIF1<0時(shí)，SKIPIF1<0≤tanα<1，即SKIPIF1<0≤k<1；當(dāng)SKIPIF1<0≤α<π時(shí)，SKIPIF1<0≤tanα<0，即SKIPIF1<0≤k＜0.∴k∈SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】本題考查直線(xiàn)的傾斜角與斜率，解決本題的關(guān)鍵是直線(xiàn)傾斜角的正切值為直線(xiàn)的斜率.16．（2023·全國(guó)·高三專(zhuān)題練習(xí)）一條直線(xiàn)經(jīng)過(guò)點(diǎn)SKIPIF1<0，并且它的傾斜角等于直線(xiàn)SKIPIF1<0的傾斜角的2倍，則這條直線(xiàn)的一般式方程是.【答案】SKIPIF1<0【解析】設(shè)直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，則所求直線(xiàn)的傾斜角為SKIPIF1<0，求出求直線(xiàn)的斜率為SKIPIF1<0，利用點(diǎn)斜式求出直線(xiàn)方程，化為一般式即可.【詳解】設(shè)直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，則所求直線(xiàn)的傾斜角為SKIPIF1<0，由SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，即所求直線(xiàn)的斜率為SKIPIF1<0，又該直線(xiàn)經(jīng)過(guò)點(diǎn)SKIPIF1<0，故所求直線(xiàn)方程為：SKIPIF1<0，即SKIPIF1<0.故答案為：SKIPIF1<0.17．（2023·上海黃浦·上海市敬業(yè)中學(xué)校考三模）若直線(xiàn)SKIPIF1<0的傾斜角為α，則sin2α的值為.【答案】SKIPIF1<0【分析】根據(jù)直線(xiàn)斜率為傾斜角的正切值，結(jié)合三角恒等變換公式即可求解.【詳解】由題可知，SKIPIF1<0，則SKIPIF1<0.故答案為：SKIPIF1<0.18．（2023·高三課時(shí)練習(xí)）直線(xiàn)SKIPIF1<0和直線(xiàn)SKIPIF1<0的夾角大小是【答案】SKIPIF1<0【分析】由題意分別求出兩條直線(xiàn)的傾斜角，即可得答案．【詳解】SKIPIF1<0直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，傾斜角為SKIPIF1<0，∴直線(xiàn)SKIPIF1<0和直線(xiàn)SKIPIF1<0的夾角大小為SKIPIF1<0，故答案為：SKIPIF1<0．【點(diǎn)睛】本題考查直線(xiàn)的傾斜角和斜率，考查運(yùn)算求解能力，屬于基礎(chǔ)題.19．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0與連接SKIPIF1<0的線(xiàn)段總有公共點(diǎn)，則SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【分析】畫(huà)出圖形，由圖可得，要使直線(xiàn)與線(xiàn)段SKIPIF1<0總有公共點(diǎn)，需滿(mǎn)足SKIPIF1<0或SKIPIF1<0，從而可求得答案【詳解】得直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，且過(guò)定點(diǎn)SKIPIF1<0，則由圖可得，要使直線(xiàn)與線(xiàn)段SKIPIF1<0總有公共點(diǎn)，需滿(mǎn)足SKIPIF1<0或SKIPIF1<0，SKIPIF1<0，SKIPIF1<0或SKIPIF1<0，SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<020．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象有且只有一個(gè)公共點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【分析】根據(jù)題意畫(huà)出圖象，結(jié)合圖象即可求解結(jié)論．【詳解】函數(shù)SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，如圖：結(jié)合圖象可得：SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0，故答案為：SKIPIF1<0，SKIPIF1<0．【B組

在綜合中考查能力】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，直線(xiàn)SKIPIF1<0的傾斜角為直線(xiàn)SKIPIF1<0的傾斜角的一半，則直線(xiàn)SKIPIF1<0的斜率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．不存在【答案】C【分析】根據(jù)斜率與傾斜角的關(guān)系，結(jié)合正切的二倍角公式，可得答案.【詳解】由直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，設(shè)其傾斜角為SKIPIF1<0，則SKIPIF1<0，由直線(xiàn)SKIPIF1<0的傾斜角為直線(xiàn)SKIPIF1<0的傾斜角的一半，設(shè)直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，由傾斜角的取值范圍為SKIPIF1<0，則SKIPIF1<0，故直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0.故選：C.2．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0的斜率的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】將直線(xiàn)的一般方程轉(zhuǎn)化為直線(xiàn)的斜截式方程，根據(jù)SKIPIF1<0的范圍求出SKIPIF1<0的范圍，進(jìn)而求出SKIPIF1<0范圍即可求解.【詳解】當(dāng)SKIPIF1<0時(shí)，直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0時(shí)，SKIPIF1<0或SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)，直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0.所以直線(xiàn)SKIPIF1<0的斜率的取值范圍為SKIPIF1<0.綜上所述，直線(xiàn)SKIPIF1<0的斜率的取值范圍為SKIPIF1<0.故選：A.3．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知過(guò)定點(diǎn)直線(xiàn)SKIPIF1<0在兩坐標(biāo)軸上的截距都是正值，且截距之和最小，則直線(xiàn)的方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題意可知，SKIPIF1<0，求出直線(xiàn)SKIPIF1<0與兩坐標(biāo)軸的交點(diǎn)SKIPIF1<0，SKIPIF1<0，再由均值不等式即可求出截距之和的最小值，即可求出直線(xiàn)方程.【詳解】直線(xiàn)SKIPIF1<0可變?yōu)镾KIPIF1<0，所以過(guò)定點(diǎn)SKIPIF1<0，又因?yàn)橹本€(xiàn)SKIPIF1<0在兩坐標(biāo)軸上的截距都是正值，可知SKIPIF1<0，令SKIPIF1<0，所以直線(xiàn)與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0，令SKIPIF1<0，所以直線(xiàn)與SKIPIF1<0軸的交點(diǎn)為SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)取等，所以此時(shí)直線(xiàn)為：SKIPIF1<0.故選：C.4．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若過(guò)點(diǎn)P(1－a,1＋a)和Q(3,2a)的直線(xiàn)的傾斜角為鈍角，則實(shí)數(shù)a的取值范圍是()A．(－2,1) B．(－1,2)C．(－∞，0) D．(－∞，－2)∪(1，＋∞)【答案】A【詳解】∵過(guò)點(diǎn)SKIPIF1<0和SKIPIF1<0的直線(xiàn)的傾斜角為鈍角∴直線(xiàn)的斜率小于0，即SKIPIF1<0.∴SKIPIF1<0∴SKIPIF1<0故選A.5．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若過(guò)點(diǎn)SKIPIF1<0的直線(xiàn)與以點(diǎn)SKIPIF1<0為端點(diǎn)的線(xiàn)段相交，則直線(xiàn)的傾斜角取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】先在直角坐標(biāo)系中作出SKIPIF1<0三點(diǎn)，再求出SKIPIF1<0的斜率，進(jìn)而求出對(duì)應(yīng)的傾斜角，結(jié)合圖象可知直線(xiàn)的傾斜角的取值范圍.【詳解】如圖所示，設(shè)SKIPIF1<0的傾斜角為SKIPIF1<0，SKIPIF1<0的傾斜角為SKIPIF1<0，則所求直線(xiàn)的傾斜角的取值范圍為SKIPIF1<0，易得SKIPIF1<0，SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以所求直線(xiàn)的傾斜角的取值范圍為SKIPIF1<0.故選：A.

6．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若SKIPIF1<0，則經(jīng)過(guò)兩點(diǎn)SKIPIF1<0，SKIPIF1<0的直線(xiàn)的傾斜角為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】先利用兩點(diǎn)的坐標(biāo)求出直線(xiàn)的斜率，再利用誘導(dǎo)公式驗(yàn)證各選項(xiàng).【詳解】由題意，得該直線(xiàn)的斜率為SKIPIF1<0，且SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以該直線(xiàn)的傾斜角為SKIPIF1<0.故選：B．7．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0和過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0相交于點(diǎn)SKIPIF1<0不重合），則SKIPIF1<0面積的最大值是（

）A．SKIPIF1<0 B．5 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由題意結(jié)合直線(xiàn)位置關(guān)系的判斷可得兩直線(xiàn)互相垂直，由直線(xiàn)過(guò)定點(diǎn)可得定點(diǎn)SKIPIF1<0與定點(diǎn)SKIPIF1<0，進(jìn)而可得SKIPIF1<0，再利用基本不等式及三角形面積公式即得．【詳解】由題意直線(xiàn)SKIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，直線(xiàn)SKIPIF1<0可變?yōu)镾KIPIF1<0，所以該直線(xiàn)過(guò)定點(diǎn)SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0互相垂直，所以SKIPIF1<0，所以SKIPIF1<0即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),所以，SKIPIF1<0，即SKIPIF1<0面積的最大值是SKIPIF1<0.故選：D.8．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三個(gè)數(shù)成等差數(shù)列，直線(xiàn)SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0，且SKIPIF1<0在直線(xiàn)SKIPIF1<0上，其中SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．2 D．4【答案】B【分析】先由等差數(shù)列求得SKIPIF1<0，再由SKIPIF1<0求出定點(diǎn)SKIPIF1<0坐標(biāo)，代入直線(xiàn)SKIPIF1<0得SKIPIF1<0，由SKIPIF1<0結(jié)合基本不等式即可求解.【詳解】易知SKIPIF1<0，則SKIPIF1<0，整理得SKIPIF1<0，由SKIPIF1<0解得SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)取等，故SKIPIF1<0的最小值為SKIPIF1<0.故選：B.9．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0交于點(diǎn)SKIPIF1<0，則SKIPIF1<0到坐標(biāo)原點(diǎn)距離的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】?jī)芍本€(xiàn)均過(guò)定點(diǎn)且垂直，則交點(diǎn)P在以?xún)啥c(diǎn)為直徑的圓上，由數(shù)形結(jié)合可求最值.【詳解】?jī)芍本€(xiàn)滿(mǎn)足SKIPIF1<0，所以?xún)芍本€(xiàn)垂直，由SKIPIF1<0得SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0，故交點(diǎn)P在以AB為直徑的圓C上，其中SKIPIF1<0，如圖所示，則線(xiàn)段OP的最大值為SKIPIF1<0.故選：B.10．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)l過(guò)點(diǎn)SKIPIF1<0，且分別交兩直線(xiàn)SKIPIF1<0于x軸上方的SKIPIF1<0兩點(diǎn)，O點(diǎn)為坐標(biāo)原點(diǎn)，則SKIPIF1<0面積的最小值為（

）A．8 B．9 C．SKIPIF1<0 D．20【答案】A【分析】判斷直線(xiàn)斜率存在并設(shè)直線(xiàn)l的方程為SKIPIF1<0，求出SKIPIF1<0兩點(diǎn)的橫坐標(biāo)，表示出三角形的面積，并化簡(jiǎn)，結(jié)合基本不等式即可求得答案.【詳解】由題意知直線(xiàn)l的斜率一定存在，斜率設(shè)為k，則直線(xiàn)l的方程為SKIPIF1<0，分別與SKIPIF1<0聯(lián)立可得SKIPIF1<0兩點(diǎn)的橫坐標(biāo)：SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0兩點(diǎn)都在x軸的上方，故SKIPIF1<0，故SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0面積的最小值為8，故選：A．11．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0：SKIPIF1<0的傾斜角為SKIPIF1<0，直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，且直線(xiàn)SKIPIF1<0在SKIPIF1<0軸上的截距為3，則直線(xiàn)SKIPIF1<0的一般式方程為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)正切二倍角公式，斜截式方程求解即可.【詳解】解：∵直線(xiàn)SKIPIF1<0：SKIPIF1<0的傾斜角為SKIPIF1<0，斜率為SKIPIF1<0，∴SKIPIF1<0，∵直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0，∴斜率為SKIPIF1<0，∴SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0.故選：B．二、多選題12．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)l過(guò)點(diǎn)SKIPIF1<0且斜率為k，若直線(xiàn)l與線(xiàn)段AB有公共點(diǎn)，SKIPIF1<0，SKIPIF1<0，則k可以取（

）A．－8 B．－5 C．3 D．4【答案】AD【分析】根據(jù)題意，做出圖形，分析直線(xiàn)斜率可知SKIPIF1<0，再利用斜率公式求解SKIPIF1<0，SKIPIF1<0即可.【詳解】解：由于直線(xiàn)l過(guò)點(diǎn)SKIPIF1<0且斜率為k，與連接兩點(diǎn)SKIPIF1<0，SKIPIF1<0的線(xiàn)段有公共點(diǎn)，則SKIPIF1<0，SKIPIF1<0，由圖可知，SKIPIF1<0時(shí)，直線(xiàn)與線(xiàn)段有交點(diǎn)，根據(jù)選項(xiàng)，可知AD符合．故選：AD．13．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)過(guò)點(diǎn)SKIPIF1<0，且在兩坐標(biāo)軸上截距的絕對(duì)值相等，則直線(xiàn)SKIPIF1<0方程可能為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【分析】討論直線(xiàn)過(guò)原點(diǎn)時(shí)和直線(xiàn)不過(guò)原點(diǎn)時(shí)，分別求出對(duì)應(yīng)的直線(xiàn)方程即可．【詳解】當(dāng)直線(xiàn)經(jīng)過(guò)原點(diǎn)時(shí)，斜率為SKIPIF1<0，所求的直線(xiàn)方程為y=2x，即SKIPIF1<0；當(dāng)直線(xiàn)不過(guò)原點(diǎn)時(shí)，設(shè)所求的直線(xiàn)方程為x±y=k，把點(diǎn)A（1，2）代入可得1-2=k，或1+2=k，求得k=-1，或k=3，故所求的直線(xiàn)方程為SKIPIF1<0，或SKIPIF1<0；綜上知，所求的直線(xiàn)方程為SKIPIF1<0、SKIPIF1<0，或SKIPIF1<0．故選：ABC．【點(diǎn)睛】本題考查了利用分類(lèi)討論思想求直線(xiàn)方程的問(wèn)題，是基礎(chǔ)題．14．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)SKIPIF1<0，則下列表述正確的是（

）A．當(dāng)SKIPIF1<0時(shí)，直線(xiàn)的傾斜角為SKIPIF1<0B．當(dāng)實(shí)數(shù)SKIPIF1<0變化時(shí)，直線(xiàn)SKIPIF1<0恒過(guò)點(diǎn)SKIPIF1<0C．當(dāng)直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0平行時(shí)，則兩條直線(xiàn)的距離為1D．直線(xiàn)SKIPIF1<0與兩坐標(biāo)軸正半軸圍成的三角形面積的最小值為4【答案】ABD【分析】A選項(xiàng)，可求出直線(xiàn)斜率，即可判斷選項(xiàng)正誤；B選項(xiàng)，將直線(xiàn)方程整理為SKIPIF1<0，由此可得直線(xiàn)所過(guò)定點(diǎn)；C選項(xiàng)，由題可得SKIPIF1<0，后由平行直線(xiàn)距離公式可判斷選項(xiàng)；D選項(xiàng)，分別令SKIPIF1<0，可得直線(xiàn)與SKIPIF1<0軸，x軸交點(diǎn)為SKIPIF1<0，SKIPIF1<0.則圍成三角形面積為SKIPIF1<0，后由基本不等式可判斷選項(xiàng).【詳解】A選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，直線(xiàn)方程為SKIPIF1<0，可得直線(xiàn)斜率為1，則傾斜角為SKIPIF1<0，故A正確；B選項(xiàng)，由題可得SKIPIF1<0，則直線(xiàn)過(guò)定點(diǎn)SKIPIF1<0，故B正確；C選項(xiàng)，因直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0平行，則SKIPIF1<0，則直線(xiàn)方程為：SKIPIF1<0，即SKIPIF1<0.則SKIPIF1<0與直線(xiàn)SKIPIF1<0之間的距離為SKIPIF1<0，故C錯(cuò)誤；D選項(xiàng)，分別令SKIPIF1<0，可得直線(xiàn)與SKIPIF1<0軸，x軸交點(diǎn)為SKIPIF1<0，SKIPIF1<0.又交點(diǎn)在兩坐標(biāo)軸正半軸，則SKIPIF1<0.故圍成三角形面積為SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào).即面積最小值為4，故D正確.故選：ABD.15．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知直線(xiàn)l經(jīng)過(guò)點(diǎn)SKIPIF1<0和SKIPIF1<0，則下列說(shuō)法正確的是（

）A．直線(xiàn)l在兩坐標(biāo)軸上的截距相等B．直線(xiàn)l的斜率為1C．原點(diǎn)到直線(xiàn)l的距離為SKIPIF1<0D．直線(xiàn)l的一個(gè)方向向量為SKIPIF1<0【答案】BC【分析】由直線(xiàn)l經(jīng)過(guò)的兩點(diǎn)坐標(biāo)，可以求出直線(xiàn)的斜率、直線(xiàn)的方程，利用直線(xiàn)的方程判斷選項(xiàng)的正誤.【詳解】直線(xiàn)l經(jīng)過(guò)點(diǎn)SKIPIF1<0和SKIPIF1<0，所以直線(xiàn)的斜率SKIPIF1<0，故B正確；易得直線(xiàn)的方程為SKIPIF1<0，即SKIPIF1<0，令SKIPIF1<0，得SKIPIF1<0，即縱截距為1，令SKIPIF1<0，得SKIPIF1<0，即橫截距為SKIPIF1<0，故A錯(cuò)誤；原點(diǎn)到直線(xiàn)l的距離SKIPIF1<0，故C正確；因?yàn)镾KIPIF1<0，所以SKIPIF1<0不是直線(xiàn)l的一個(gè)方向向量，故D錯(cuò)誤；故選：BC.16．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0的方程為：SKIPIF1<0，則（

）A．直線(xiàn)SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0B．直線(xiàn)SKIPIF1<0斜率必定存在C．SKIPIF1<0時(shí)直線(xiàn)SKIPIF1<0的傾斜角為SKIPIF1<0D．SKIPIF1<0時(shí)直線(xiàn)SKIPIF1<0與兩坐標(biāo)軸圍成的三角形面積為SKIPIF1<0【答案】AD【分析】利用直線(xiàn)系方程可判斷A，判斷直線(xiàn)的斜率可判斷B，求直線(xiàn)的傾斜角可判斷C，求解三角形的面積可判斷D.【詳解】對(duì)于A，由直線(xiàn)方程知：恒過(guò)定點(diǎn)SKIPIF1<0，故正確；對(duì)于B，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，直線(xiàn)斜率不存在，故錯(cuò)誤；對(duì)于C，SKIPIF1<0時(shí)有SKIPIF1<0，設(shè)傾斜角為SKIPIF1<0，即SKIPIF1<0，則傾斜角為SKIPIF1<0，故錯(cuò)誤；對(duì)于D，SKIPIF1<0時(shí)，直線(xiàn)SKIPIF1<0，則x、y軸交點(diǎn)分別為SKIPIF1<0，所以直線(xiàn)與兩坐標(biāo)軸圍成的三角形面積為SKIPIF1<0，故正確；故選：AD.17．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)直線(xiàn)l的方程為SKIPIF1<0．下列說(shuō)法正確的是（

）A．當(dāng)SKIPIF1<0時(shí)，l不經(jīng)過(guò)第二象限B．直線(xiàn)恒過(guò)定點(diǎn)SKIPIF1<0C．不論a為何值，直線(xiàn)恒過(guò)第四象限D(zhuǎn)．直線(xiàn)的傾斜角不可能是90°【答案】ACD【分析】將直線(xiàn)l變形為斜截式，由l不經(jīng)過(guò)第二象限，列出關(guān)于a的不等關(guān)系，求解即可判斷選項(xiàng)A，將點(diǎn)代入方程即可判斷選項(xiàng)B，由直線(xiàn)恒過(guò)定點(diǎn)SKIPIF1<0，即可判斷選項(xiàng)C，由斜率與傾斜角的關(guān)系，即可判斷選項(xiàng)D.【詳解】對(duì)于A，將l的方程化為SKIPIF1<0，欲使l不經(jīng)過(guò)第二象限，當(dāng)且僅當(dāng)SKIPIF1<0或SKIPIF1<0成立，所以SKIPIF1<0，故A正確；對(duì)于B，點(diǎn)SKIPIF1<0代入直線(xiàn)方程不成立，B不正確；對(duì)于C，因?yàn)橹本€(xiàn)恒過(guò)第四象限內(nèi)的點(diǎn)SKIPIF1<0，所以不論a為何值，直線(xiàn)恒過(guò)第四象限，C正確；對(duì)于D，直線(xiàn)的斜率始終存在，為SKIPIF1<0，所以?xún)A斜角不可能等于90°，D正確．故選：ACD三、填空題18．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0不經(jīng)過(guò)第二象限，則實(shí)數(shù)SKIPIF1<0的取值范圍為．【答案】SKIPIF1<0【分析】根據(jù)直線(xiàn)的斜率和在SKIPIF1<0軸上的截距建立不等式組求解即可.【詳解】由直線(xiàn)不過(guò)第二象限需滿(mǎn)足SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<019．（2023·全國(guó)·高三專(zhuān)題練習(xí)）在平面直角坐標(biāo)系中，已知SKIPIF1<0兩點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn)，則SKIPIF1<0的平分線(xiàn)所在直線(xiàn)的方程為.【答案】SKIPIF1<0【分析】設(shè)SKIPIF1<0的平分線(xiàn)的傾斜角為SKIPIF1<0，根據(jù)斜率公式結(jié)合SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0的范圍即可求解.【詳解】由題意，可設(shè)SKIPIF1<0的平分線(xiàn)的傾斜角為SKIPIF1<0，如圖，則SKIPIF1<0，即SKIPIF1<0.則SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0的平分線(xiàn)所在直線(xiàn)的方程為SKIPIF1<0，故答案為：SKIPIF1<020．（2023·全國(guó)·高三專(zhuān)題練習(xí)）過(guò)點(diǎn)SKIPIF1<0作直線(xiàn)SKIPIF1<0分別交SKIPIF1<0軸、SKIPIF1<0軸的正半軸于SKIPIF1<0、SKIPIF1<0兩點(diǎn)，則使SKIPIF1<0的值最小時(shí)直線(xiàn)SKIPIF1<0的方程為．【答案】SKIPIF1<0【分析】利用三角函數(shù)的定義求得SKIPIF1<0關(guān)于SKIPIF1<0的表示，再利用倍角公式與正弦函數(shù)的性質(zhì)即可得解.【詳解】如圖所示：設(shè)SKIPIF1<0，SKIPIF1<0，

SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0取最小值，此時(shí)，直線(xiàn)的傾斜角為SKIPIF1<0，斜率為SKIPIF1<0，∴直線(xiàn)的方程為SKIPIF1<0，即SKIPIF1<0．故答案為：SKIPIF1<0.21．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，則直線(xiàn)SKIPIF1<0恒過(guò)定點(diǎn)．【答案】SKIPIF1<0【分析】根據(jù)直線(xiàn)SKIPIF1<0恒過(guò)定點(diǎn)，求其關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱(chēng)點(diǎn)，即可求解.【詳解】因?yàn)镾KIPIF1<0過(guò)定點(diǎn)SKIPIF1<0，而SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱(chēng)點(diǎn)為SKIPIF1<0，又直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng)，所以直線(xiàn)SKIPIF1<0恒過(guò)定點(diǎn)SKIPIF1<0.【點(diǎn)睛】本題主要考查了直線(xiàn)系過(guò)定點(diǎn)，直線(xiàn)關(guān)于點(diǎn)對(duì)稱(chēng)，點(diǎn)關(guān)于點(diǎn)對(duì)稱(chēng)問(wèn)題，屬于中檔題.22．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0過(guò)點(diǎn)P（1，4），分別交x軸的正半軸和y軸的正半軸于點(diǎn)A，B兩點(diǎn)，O為坐標(biāo)原點(diǎn)，當(dāng)SKIPIF1<0最小時(shí)，SKIPIF1<0的方程為.【答案】SKIPIF1<0【分析】由題意知直線(xiàn)斜率存在，設(shè)直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0求出與坐標(biāo)軸交點(diǎn)計(jì)算SKIPIF1<0，由均值不等式求最值.【詳解】經(jīng)檢驗(yàn)直線(xiàn)SKIPIF1<0的斜率存在，且斜率為負(fù)，設(shè)直線(xiàn)SKIPIF1<0的斜率為SKIPIF1<0，則直線(xiàn)SKIPIF1<0的方程為SKIPIF1<0，令y=0得SKIPIF1<0，令x=0得SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0取得最小值.此時(shí)SKIPIF1<0的方程為SKIPIF1<0.故答案為：SKIPIF1<023．（2023·全國(guó)·高三專(zhuān)題練習(xí)）直線(xiàn)SKIPIF1<0的傾斜角SKIPIF1<0的取值范圍是.【答案】SKIPIF1<0【分析】分別討論SKIPIF1<0的取值，得到斜率不存在時(shí)SKIPIF1<0，以及斜率存在時(shí)SKIPIF1<0的范圍，再利用傾斜角與斜率的關(guān)系，即可求解.【詳解】若SKIPIF1<0，則直線(xiàn)方程為SKIPIF1<0，即傾斜角SKIPIF1<0；若SKIPIF1<0，則直線(xiàn)方程為SKIPIF1<0，即SKIPIF1<0，∵SKIPIF1<0，∴SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0，解得SKIPIF1<0綜上可得SKIPIF1<0.故答案為：SKIPIF1<024．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知點(diǎn)A（2，-1），B（3，m），若SKIPIF1<0，則直線(xiàn)AB的傾斜角的取值范圍為_(kāi)_________．【答案】SKIPIF1<0【分析】設(shè)直線(xiàn)AB的傾斜角為α，由點(diǎn)A，B的坐標(biāo)求出直線(xiàn)AB的斜率k，結(jié)合m的范圍可得k的斜率，即tanα的范圍，再利用正切函數(shù)的性質(zhì)即可求出α的取值范圍．【詳解】設(shè)直線(xiàn)AB的傾斜角為α，∵點(diǎn)A（2，-1），B（3，m），∴直線(xiàn)AB的斜率SKIPIF1<0，又∵SKIPIF1<0，∴SKIPIF1<0，即k的取值范圍為SKIPIF1<0，即SKIPIF1<0，又∵α∈[0，π），∴SKIPIF1<0，故答案為：SKIPIF1<0．25．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)SKIPIF1<0，過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0和過(guò)定點(diǎn)SKIPIF1<0的動(dòng)直線(xiàn)SKIPIF1<0交于點(diǎn)SKIPIF1<0，則SKIPIF1<0的最大值.【答案】SKIPIF1<0【分析】根據(jù)兩直線(xiàn)的方程可求得定點(diǎn)SKIPIF1<0、SKIPIF1<0的坐標(biāo)，以及兩直線(xiàn)垂直，進(jìn)而可得SKIPIF1<0，再結(jié)合SKIPIF1<0即可求解.【詳解】由SKIPIF1<0可知SKIPIF1<0，所以該直線(xiàn)過(guò)定點(diǎn)SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以該直線(xiàn)過(guò)定點(diǎn)SKIPIF1<0，因?yàn)橹本€(xiàn)SKIPIF1<0與SKIPIF1<0垂直，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，即SKIPIF1<0，解得：SKIPIF1<0，所以SKIPIF1<0的最大值為SKIPIF1<0，故答案為：SKIPIF1<0.26．（2023·全國(guó)·高三專(zhuān)題練習(xí)）如圖，射線(xiàn)OA，OB分別與x軸正半軸成SKIPIF1<0和SKIPIF1<0角，過(guò)點(diǎn)SKIPIF1<0作直線(xiàn)AB分別交OA，OB于A，B兩點(diǎn)，當(dāng)AB的中點(diǎn)C恰好落在直線(xiàn)SKIPIF1<0上時(shí)，則直線(xiàn)AB的方程是.【答案】SKIPIF1<0【分析】由題意求出直線(xiàn)SKIPIF1<0的方程，設(shè)SKIPIF1<0得到AB的中點(diǎn)的坐標(biāo)，由A，P，B三點(diǎn)共線(xiàn)求出SKIPIF1<0，得到直線(xiàn)SKIPIF1<0的斜率，再利用直線(xiàn)的點(diǎn)斜式方程可得答案.【詳解】由題意可得SKIPIF1<0，SKIPIF1<0，所以直線(xiàn)SKIPIF1<0，設(shè)SKIPIF1<0，所以AB的中點(diǎn)CSKIPIF1<0.由點(diǎn)C在直線(xiàn)SKIPIF1<0上，且A，P，B三點(diǎn)共線(xiàn)得SKIPIF1<0解得SKIPIF1<0，所以SKIPIF1<0,又SKIPIF1<0，所以SKIPIF1<0＝SKIPIF1<0，所以SKIPIF1<0，即直線(xiàn)AB的方程為SKIPIF1<0.【C組

在創(chuàng)新中考查思維】一、單選題1．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若直線(xiàn)SKIPIF1<0與直線(xiàn)SKIPIF1<0相交，且交點(diǎn)在第一象限，則直線(xiàn)SKIPIF1<0的傾斜角的取值范圍是A．SKIP