新高考數(shù)學(xué)一輪復(fù)習(xí)精講精練10.1 直線方程（基礎(chǔ)版）（解析版）

10.1直線方程（精講）（基礎(chǔ)版）思維導(dǎo)圖思維導(dǎo)圖考點(diǎn)呈現(xiàn)考點(diǎn)呈現(xiàn)例題剖析例題剖析考點(diǎn)一直線的傾斜角與斜率【例1-1】（2021廣安期末）直線SKIPIF1<0的傾斜角為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由已知得SKIPIF1<0，故直線斜率SKIPIF1<0由于傾斜的范圍是SKIPIF1<0，則傾斜角為SKIPIF1<0.故答案為：B.【例1-2】（2022梅州期末）已知SKIPIF1<0，且SKIPIF1<0三點(diǎn)共線，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0三點(diǎn)共線，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0。故答案為：A.【例1-3】（2022達(dá)州期末）已知SKIPIF1<0，SKIPIF1<0，過點(diǎn)SKIPIF1<0且斜率為SKIPIF1<0的直線l與線段AB有公共點(diǎn)，則SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】因?yàn)檫^點(diǎn)SKIPIF1<0且斜率為SKIPIF1<0的直線l與線段AB有公共點(diǎn)，所以由圖可知，SKIPIF1<0或SKIPIF1<0，因?yàn)镾KIPIF1<0或SKIPIF1<0，所以SKIPIF1<0或SKIPIF1<0，故答案為：D【一隅三反】1．（2022浙江期中）直線SKIPIF1<0的傾斜角為（）A．30° B．60° C．120° D．150°【答案】C【解析】直線SKIPIF1<0的斜率SKIPIF1<0設(shè)其傾斜角為SKIPIF1<0，故可得SKIPIF1<0，又SKIPIF1<0，故SKIPIF1<0.故答案為：C.2．（2022·楊浦二模）橢圓C：SKIPIF1<0的左、右頂點(diǎn)分別為SKIPIF1<0，SKIPIF1<0，點(diǎn)P在C上（P不與SKIPIF1<0，SKIPIF1<0重合）且直線SKIPIF1<0的斜率的取值范圍是SKIPIF1<0，那么直線SKIPIF1<0斜率的取值范圍是（）A．[SKIPIF1<0，SKIPIF1<0] B．[SKIPIF1<0，SKIPIF1<0] C．[SKIPIF1<0，1] D．[SKIPIF1<0，1]【答案】B【解析】設(shè)P點(diǎn)坐標(biāo)為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，于是SKIPIF1<0，故SKIPIF1<0.∵SKIPIF1<0∴SKIPIF1<0.故答案為：B.3．（2022達(dá)州期末）點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)辄c(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上，所以SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；故設(shè)SKIPIF1<0而SKIPIF1<0可看作函數(shù)SKIPIF1<0的圖象上的點(diǎn)與點(diǎn)SKIPIF1<0（-1，-2）連線的斜率，故SKIPIF1<0時(shí)，SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0故答案為：B.考點(diǎn)二直線的方程【例2-1】（2021嘉興期末）過點(diǎn)SKIPIF1<0且垂直于直線SKIPIF1<0的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】設(shè)所求的直線方程為SKIPIF1<0，SKIPIF1<0代入方程解得SKIPIF1<0，所求的直線方程為SKIPIF1<0.故答案為：D.【例2-2】（2022漢中期中）直線SKIPIF1<0在y軸上的截距為（）A．-1 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，可得SKIPIF1<0，則直線SKIPIF1<0在SKIPIF1<0軸上的截距為-1。故答案為：A【例2-3】（2021深圳期末）將一張坐標(biāo)紙折疊一次，使點(diǎn)SKIPIF1<0與SKIPIF1<0重合，求折痕所在直線是（）．A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0的中點(diǎn)坐標(biāo)為SKIPIF1<0，又SKIPIF1<0，所以折痕所在直線的斜率為1，故折痕所在直線是SKIPIF1<0，即SKIPIF1<0。故答案為：D【一隅三反】1．（2021東城期末）已知SKIPIF1<0的三個(gè)頂點(diǎn)是SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則邊SKIPIF1<0上的高所在的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】因?yàn)镾KIPIF1<0，所以邊SKIPIF1<0上的高所在的直線的斜率為SKIPIF1<0，所以邊SKIPIF1<0上的高所在的直線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：B.2．（2022·濟(jì)南模擬）過SKIPIF1<0與SKIPIF1<0的交點(diǎn)，且平行于向量SKIPIF1<0的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由SKIPIF1<0，得SKIPIF1<0，所以交點(diǎn)坐標(biāo)為SKIPIF1<0，又因?yàn)橹本€平行于向量SKIPIF1<0，所以所求直線方程為SKIPIF1<0，即SKIPIF1<0.故答案為：C.3．（2021蘭溪期中）過點(diǎn)SKIPIF1<0和SKIPIF1<0的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】設(shè)直線的方程為SKIPIF1<0又因?yàn)橹本€經(jīng)過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，所以直線的方程為SKIPIF1<0所以直線方程為SKIPIF1<0。故答案為：C考點(diǎn)三直線的位置關(guān)系【例3-1】（2021廣安期末）“SKIPIF1<0”是“直線SKIPIF1<0與直線SKIPIF1<0垂直”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0所以SKIPIF1<0，反之，則不然所以“SKIPIF1<0”是“直線SKIPIF1<0與直線SKIPIF1<0垂直”的充分不必要條件.故答案為：A【例3-2】（2022廣東月考）若直線SKIPIF1<0與直線SKIPIF1<0平行，則m=（）A．4 B．-4 C．1 D．-1【答案】A【解析】因?yàn)橹本€SKIPIF1<0與直線SKIPIF1<0平行，所以SKIPIF1<0，解得SKIPIF1<0．故答案為：A【一隅三反】1．（2022浙江月考）已知直線SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的值是（）A．-1 B．2 C．2或-1 D．-2或1【答案】A【解析】由兩直線平行，可知SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)兩直線平行；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，此時(shí)兩直線重合，不滿足題意舍去．故答案為：A.2．（2022江蘇）若SKIPIF1<0，則“SKIPIF1<0”是“直線SKIPIF1<0和直線SKIPIF1<0平行”的（）A．充分不必要條件 B．充要條件C．必要不充分條件 D．既不充分又不必要條件【答案】C【解析】由直線ax+y-1=0和直線x+by-1=0平行，可得ab=1.

反之不成立，例如a=b=1時(shí)，兩條直線都為x+y-1=0，所以兩條直線重合.

ab=1是“直線ax+y-1=0和直線x+by-1=0平行”的必要不充分條件.

故選C.

3（2022上海）．“SKIPIF1<0”是“直線SKIPIF1<0與SKIPIF1<0平行”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分又不必要條件【答案】A【解析】充分性：當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0即為：SKIPIF1<0與SKIPIF1<0，所以兩直線平行.故充分性滿足；必要性：直線SKIPIF1<0與SKIPIF1<0平行，則有：SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0即為：SKIPIF1<0與SKIPIF1<0，所以兩直線平行，不重合；當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0與SKIPIF1<0即為：SKIPIF1<0與SKIPIF1<0，所以兩直線平行，不重合；所以SKIPIF1<0或SKIPIF1<0.故必要性不滿足.故“SKIPIF1<0”是“直線SKIPIF1<0與SKIPIF1<0平行”的充分不必要條件.故答案為：A考點(diǎn)四直線過定點(diǎn)【例4】（2022廣東）直線SKIPIF1<0恒過定點(diǎn)（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】將SKIPIF1<0變形為：SKIPIF1<0，令SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0，故直線恒過定點(diǎn)SKIPIF1<0故答案為：A【一隅三反】1．（2022年廣西）直線SKIPIF1<0恒過一定點(diǎn)，則此定點(diǎn)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】直線SKIPIF1<0可變形為：SKIPIF1<0，由直線的點(diǎn)斜式方程可知：直線恒過定點(diǎn)SKIPIF1<0。故答案為：A2（2022山西）．直線SKIPIF1<0恒過定點(diǎn)（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0得到：SKIPIF1<0，∴直線SKIPIF1<0恒過定點(diǎn)SKIPIF1<0．故答案為：A3（）222山東0直線l：SKIPIF1<0經(jīng)過定點(diǎn)A，則A的縱坐標(biāo)為（）A．-2 B．-1 C．1 D．2【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，令x?1=0x+2y+3=0，得SKIPIF1<0。故答案為：A考點(diǎn)五三種距離【例5-1】（2022高二下·成都開學(xué)考）雙曲線為SKIPIF1<0，則它的焦點(diǎn)到漸近線的距離為（）．A．2 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A【解析】由題意得a=1，b=2，SKIPIF1<0，不妨取焦點(diǎn)為SKIPIF1<0，漸近線為2x-y=0

則所求距離為：SKIPIF1<0故答案為：A

【例5-2】（2022·涼山模擬）已知直線SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0故答案為：D【例5-3】（2022漢中期中）直線SKIPIF1<0：SKIPIF1<0與SKIPIF1<0：SKIPIF1<0之間的距離為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0平行，故SKIPIF1<0與SKIPIF1<0之間的距離為SKIPIF1<0。故答案為：B.【一隅三反】1．（2022嫩江月考）已知SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0（）．A．SKIPIF1<0 B．4 C．5 D．SKIPIF1<0【答案】C【解析】由題意得SKIPIF1<0.故答案為：C.

2．（2022·吉林模擬）已知SKIPIF1<0兩點(diǎn)到直線SKIPIF1<0的距離相等，則SKIPIF1<0（）A．2 B．SKIPIF1<0 C．2或SKIPIF1<0 D．2或SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0兩點(diǎn)到直線SKIPIF1<0的距離相等，所以有SKIPIF1<0，或SKIPIF1<0，故答案為：D3．（2021白云期末）已知點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為1，則m的值為（）A．-5或-15 B．-5或15 C．5或-15 D．5或15【答案】D【解析】點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離為1，SKIPIF1<0解得：m=15或5．故答案為：D.考點(diǎn)六對(duì)稱問題【例6-1】（2022貴州）直線y＝4x﹣5關(guān)于點(diǎn)P（2，1）對(duì)稱的直線方程是（）A．y＝4x+5 B．y＝4x﹣5 C．y＝4x﹣9 D．y＝4x+9【答案】C【解析】設(shè)直線SKIPIF1<0上的點(diǎn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱點(diǎn)的坐標(biāo)為SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，將其代入直線SKIPIF1<0中，得到SKIPIF1<0，化簡得SKIPIF1<0。故答案為：C．【例6-2】（2022西安）求直線x＋2y－1＝0關(guān)于直線x＋2y＋1＝0對(duì)稱的直線方程（）A．x＋2y－3＝0 B．x＋2y＋3＝0 C．x＋2y－2＝0 D．x＋2y＋2＝0【答案】B【解析】設(shè)對(duì)稱直線方程為SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），所以所求直線方程為SKIPIF1<0。故答案為：B【一隅三反】1（2022天津）如果SKIPIF1<0關(guān)于直線l的對(duì)稱點(diǎn)為SKIPIF1<0，則直線l的方程是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)橐阎c(diǎn)SKIPIF1<0關(guān)于直線l的對(duì)稱點(diǎn)為SKIPIF1<0，故直線l為線段SKIPIF1<0的中垂線，求得SKIPIF1<0的中點(diǎn)坐標(biāo)為SKIPIF1<0，SKIPIF1<0的斜率為SKIPIF1<0，故直線l的斜率為-3，故直線l的方程為SKIPIF1<0，即SKIPIF1<0。故答案為：A.2．（2022云南）已知直線SKIPIF1<0，直線SKIPIF1<0，則SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題知直線SKIPIF1<0與直線SKIPIF1<0交于點(diǎn)SKIPIF1<0，且點(diǎn)SKIPIF1<0在SKIPIF1<0上，設(shè)點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱的點(diǎn)的坐標(biāo)為SKIPIF1<0，則SKIPIF1<0解得SKIPIF1<0則直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0關(guān)于SKIPIF1<0對(duì)稱的直線方程為SKIPIF1<0．故答案為：D3（2022西藏）．已知直線SKIPIF1<0:SKIPIF1<0，點(diǎn)SKIPIF1<0．（1）求點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0的坐標(biāo)；（2）直線SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0對(duì)稱的直線SKIPIF1<0的方程；（3）以SKIPIF1<0為圓心，3為半徑長作圓，直線SKIPIF1<0過點(diǎn)SKIPIF1<0，且被圓SKIPIF1<0截得的弦長為SKIPIF1<0，求直線SKIPIF1<0的方程．【答案】見解析【解析】（1）解：設(shè)點(diǎn)SKIPIF1<0，則SKIPIF1<0，解得：SKIPIF1<0即點(diǎn)SKIPIF1<0關(guān)于直線SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0的坐標(biāo)為SKIPIF1<0.（2）解：設(shè)SKIPIF1<0是直線SKIPIF1<0上任意一點(diǎn)，則點(diǎn)SKIPIF1<0關(guān)于點(diǎn)SKIPIF1<0的對(duì)稱點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，所以SKIPIF1<0，即SKIPIF1<0（3）解：設(shè)圓心SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，直線SKIPIF1<0被圓SKIPIF1<0截得的弦長為SKIPIF1<0，因此SKIPIF1<0，當(dāng)直線SKIPIF1<0斜率不存在時(shí)，SKIPIF1<0不滿足條件；當(dāng)直線SKIPIF1<0斜率存在時(shí)，設(shè)其方程為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，綜上，直線SKIPIF1<0的方程為SKIPIF1<0或SKIPIF1<0．10.1直線方程（精練）（基礎(chǔ)版）題組一題組一直線的傾斜角與斜率1．（2022梅州期末）已知角SKIPIF1<0的終邊過點(diǎn)SKIPIF1<0，則SKIPIF1<0可以為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】根據(jù)題意可知角SKIPIF1<0為第四象限角，則A、B不符合題意過SKIPIF1<0作SKIPIF1<0軸，垂足為SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0結(jié)合象限角的概念可得：SKIPIF1<0可以為SKIPIF1<0故答案為：C．2．（2022福州期中）直線SKIPIF1<0的傾斜角是（）A．30° B．45° C．60° D．75°【答案】B【解析】直線SKIPIF1<0的斜率為1，傾斜角為45°，故答案為：B．3．（2021浙江期末）已知點(diǎn)A（1，-1），B（1，2），則直線AB的傾斜角為（）A．0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由題意可知，SKIPIF1<0兩點(diǎn)的橫坐標(biāo)相等，則直線AB的傾斜角為SKIPIF1<0.故選：D4．（2021寧德期末）若直線經(jīng)過兩點(diǎn)SKIPIF1<0，SKIPIF1<0且傾斜角為45°，則m的值為（）A．2 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A【解析】因直線的傾斜角為SKIPIF1<0，則此直線的斜率SKIPIF1<0，而直線過點(diǎn)SKIPIF1<0，因此，SKIPIF1<0，解得SKIPIF1<0，所以m的值為2.故答案為：A5．（2022江蘇）已知SKIPIF1<0，且SKIPIF1<0三點(diǎn)共線，則SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，因?yàn)镾KIPIF1<0三點(diǎn)共線，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0。故答案為：A.6（2022黑龍江）直線SKIPIF1<0與SKIPIF1<0的夾角為．【答案】SKIPIF1<0【解析】直線SKIPIF1<0的斜率SKIPIF1<0，即傾斜角SKIPIF1<0滿足SKIPIF1<0，直線SKIPIF1<0的斜率SKIPIF1<0，即傾斜角SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又兩直線夾角的范圍為SKIPIF1<0，所以兩直線夾角為SKIPIF1<0，故答案為：SKIPIF1<0.8．（2022·虹口）直線SKIPIF1<0與SKIPIF1<0的夾角為．【答案】SKIPIF1<0【解析】直線SKIPIF1<0的斜率SKIPIF1<0，即傾斜角SKIPIF1<0滿足SKIPIF1<0，直線SKIPIF1<0的斜率SKIPIF1<0，即傾斜角SKIPIF1<0滿足SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，又兩直線夾角的范圍為SKIPIF1<0，所以兩直線夾角為SKIPIF1<0，故答案為：SKIPIF1<0.9．（2022金山）求直線SKIPIF1<0與直線SKIPIF1<0的夾角為.【答案】SKIPIF1<0【解析】SKIPIF1<0直線SKIPIF1<0的斜率不存在，傾斜角為SKIPIF1<0，直線SKIPIF1<0的斜率為SKIPIF1<0，傾斜角為SKIPIF1<0，故直線SKIPIF1<0與直線SKIPIF1<0的夾角為SKIPIF1<0，故答案為：SKIPIF1<0．題組二題組二直線方程1．（2021樂山期中）數(shù)學(xué)家歐拉在1765年提出定理：三角形的外心、重心、垂心依次位于同一直線上，且重心到外心的距離是重心到垂心距離的一半．這條直線被后人稱為三角形的歐拉線．已知△ABC的頂點(diǎn)A(1，0)，B(0，2)，且AC＝BC，則△ABC的歐拉線的方程為（）A．4x＋2y＋3＝0 B．2x－4y＋3＝0C．x－2y＋3＝0 D．2x－y＋3＝0【答案】B【解析】因?yàn)锳C＝BC，所以歐拉線為AB的中垂線，又A(1，0)，B(0，2)，AB的中點(diǎn)為SKIPIF1<0，kAB＝－2，AB的中垂線方程為y－1＝SKIPIF1<0，即2x－4y＋3＝0.故答案為：B．2．（2021懷仁期中）過點(diǎn)SKIPIF1<0且與直線SKIPIF1<0垂直的直線方程是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】直線SKIPIF1<0的斜率為SKIPIF1<0，和該直線垂直的直線的斜率為SKIPIF1<0，又因?yàn)橹本€過點(diǎn)SKIPIF1<0，故得到直線方程為SKIPIF1<0.故答案為：B.3．（2022湖南月考）已知直線SKIPIF1<0過點(diǎn)SKIPIF1<0，SKIPIF1<0，則直線SKIPIF1<0的方程為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由直線的兩點(diǎn)式方程可得，直線l的方程為SKIPIF1<0，即SKIPIF1<0。故答案為：C．4．（2021縉云月考）經(jīng)過兩條直線SKIPIF1<0和SKIPIF1<0的交點(diǎn)，且垂直于直線SKIPIF1<0的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0，解得SKIPIF1<0，因?yàn)樗笾本€與直線SKIPIF1<0垂直，所以所求直線方程：2x+3y＋c＝0，代入點(diǎn)SKIPIF1<0可得SKIPIF1<0，所以所求直線方程為SKIPIF1<0。故答案為：D5．（2022豐臺(tái)期中）過點(diǎn)SKIPIF1<0，且橫、縱截距相等的直線方程為（）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【解析】當(dāng)直線過原點(diǎn)時(shí)，直線的斜率為SKIPIF1<0，則直線方程為SKIPIF1<0；當(dāng)直線不過原點(diǎn)時(shí)，設(shè)直線方程為SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所求的直線方程為SKIPIF1<0，綜上可知，所求直線方程為SKIPIF1<0或SKIPIF1<0。故答案為：D.6．（2022河北期中）△ABC的三個(gè)頂點(diǎn)是A（4，0），B（6，7），C（0，3），則邊BC上的高所在直線的方程為（）A．5x＋y﹣20＝0 B．3x＋2y﹣12＝0C．3x＋2y﹣19＝0 D．3x﹣2y﹣12＝0【答案】B【解析】由題意，SKIPIF1<0，所以BC上的高所在直線的斜率為SKIPIF1<0，其方程為：SKIPIF1<0.

故答案為：B.

7．（2022浦城）已知A(－1，2)，B(1，3)，C(0，－2)，點(diǎn)D使AD⊥BC，AB∥CD，則點(diǎn)D的坐標(biāo)為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】設(shè)D(x，y)，∵AD⊥BC，∴SKIPIF1<0·SKIPIF1<0＝－1，∴x＋5y－9＝0，∵AB∥CD，∴SKIPIF1<0＝SKIPIF1<0，∴x－2y－4＝0，由得x+5y?9=0x?2y?4=0，x=387y=57，故答案為：D.8．（2022沈陽月考）直線SKIPIF1<0過點(diǎn)SKIPIF1<0，與直線SKIPIF1<0垂直的直線方程為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】因?yàn)橹本€SKIPIF1<0與直線SKIPIF1<0垂直，且直線SKIPIF1<0的斜率SKIPIF1<0，所以直線SKIPIF1<0的斜率SKIPIF1<0，又因?yàn)橹本€SKIPIF1<0過點(diǎn)SKIPIF1<0，所以直線SKIPIF1<0的方程為SKIPIF1<0，即SKIPIF1<0。故答案為：A.9．（2022廣州）經(jīng)過點(diǎn)SKIPIF1<0，并且在兩坐標(biāo)軸上的截距相等的直線方程是（）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】D【解析】過點(diǎn)P(2，3)，并且在兩坐標(biāo)軸上的截距相等的直線，

則直線滿足直線過原點(diǎn)，或者直線的斜率為-1，

當(dāng)直線過原點(diǎn)，則設(shè)為y=kx，則2k=3，解得SKIPIF1<0，所以直線方程為SKIPIF1<0，即3x-2y=0；

當(dāng)直線的斜率為-1時(shí)，直線方程為y-3=(-1)?(x-2)，即x+y-5=0，

所以所求直線方程為：x+y-5=0或3x-2y=0．故選：D題組三直線的位置關(guān)系題組三直線的位置關(guān)系1．（2022大連）直線l1：2x＋3y－2＝0，l2：2x＋3y＋2＝0的位置關(guān)系是（）A．垂直 B．平行 C．相交 D．重合【答案】B【解析】由題SKIPIF1<0，SKIPIF1<0，則兩直線的斜率相等，在在SKIPIF1<0軸的截距，故兩條件直線的位置關(guān)系為平行.故答案為：B2．（2022慈溪）已知直線SKIPIF1<0，SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（）A．充分不必要條件 B．必要不充分條件C．充分必要條件 D．既不充分也不必要條件【答案】C【解析】直線SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的充要條件是SKIPIF1<0，解得SKIPIF1<0因此得到“SKIPIF1<0”是“SKIPIF1<0”的充分必要條件.故答案為：C.3．（2022青島）SKIPIF1<0是直線SKIPIF1<0和SKIPIF1<0平行的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】當(dāng)SKIPIF1<0時(shí)，直線SKIPIF1<0和SKIPIF1<0分別為：SKIPIF1<0和SKIPIF1<0，顯然，兩直線平行；當(dāng)直線SKIPIF1<0和SKIPIF1<0平行時(shí)，有SKIPIF1<0成立，解得SKIPIF1<0或SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，兩直線為SKIPIF1<0和SKIPIF1<0，顯然，兩直線不重合是平行關(guān)系；當(dāng)SKIPIF1<0時(shí)，兩直線為SKIPIF1<0和SKIPIF1<0，顯然，兩直線不重合是平行關(guān)系；由此可判斷SKIPIF1<0是直線SKIPIF1<0和SKIPIF1<0平行的充分不必要條件，故答案為：A.4．（2022四川）“SKIPIF1<0”是“直線SKIPIF1<0：SKIPIF1<0與直線SKIPIF1<0：SKIPIF1<0互相垂直”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】依題意，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，所以“SKIPIF1<0”是“直線SKIPIF1<0：SKIPIF1<0與直線SKIPIF1<0：SKIPIF1<0互相垂直”的充分不必要條件.故答案為：A5．（2022云南）“SKIPIF1<0”是“直線SKIPIF1<0與直線SKIPIF1<0垂直”的（）A．充分不必要條件 B．必要不充分條件C．充要條件 D．既不充分也不必要條件【答案】A【解析】由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0所以SKIPIF1<0，反之，則不然所以“SKIPIF1<0”是“直線SKIPIF1<0與直線SKIPIF1<0垂直”的充分不必要條件.故答案為：A6．（2022廣東）已知直線SKIPIF1<0：SKIPIF1<0.直線SKIPIF1<0：SKIPIF1<0，則下列命題正確的是（）A．若SKIPIF1<0，則SKIPIF1<0 B．若SKIPIF1<0，則SKIPIF1<0C．直線SKIPIF1<0過定點(diǎn)SKIPIF1<0 D．直線SKIPIF1<0過定點(diǎn)SKIPIF1<0【答案】BCD【解析】A.若SKIPIF1<0，則SKIPIF1<0或SKIPIF1<0，經(jīng)檢驗(yàn)此時(shí)兩直線平行，所以該選項(xiàng)錯(cuò)誤；B.若SKIPIF1<0，則SKIPIF1<0，所以該選項(xiàng)正確；C.直線SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，無論SKIPIF1<0取何值，SKIPIF1<0恒成立，所以此時(shí)直線SKIPIF1<0過定點(diǎn)SKIPIF1<0，所以該選項(xiàng)正確；D.直線SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，無論SKIPIF1<0取何值，SKIPIF1<0恒成立，所以直線SKIPIF1<0過定點(diǎn)SKIPIF1<0，所以該選項(xiàng)正確.故答案為：BCD7．（2022云南）若方程組SKIPIF1<0無解，則實(shí)數(shù)SKIPIF1<0．【答案】±2【解析】因?yàn)榉匠探MSKIPIF1<0無解，所以兩直線平行，可得SKIPIF1<0.題組四題組四直線過定點(diǎn)1．（2022天津）直線SKIPIF1<0恒過定點(diǎn)為.【答案】SKIPIF1<0【解析】直線方程可化為SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，所以直線過定點(diǎn)SKIPIF1<0。故答案為：SKIPIF1<0。2．（2022·安徽）直線SKIPIF1<0經(jīng)過的定點(diǎn)坐標(biāo)是___________.【答案】SKIPIF1<0【解析】把直線l的方程改寫成：SKIPIF1<0，令SKIPIF1<0，解得：SKIPIF1<0，所以直線l總過定點(diǎn)SKIPIF1<0.故答案為：（1,1）3．（2021·重慶市）直線SKIPIF1<0：SKIPIF1<0恒過的定點(diǎn)坐標(biāo)為____________.【答案】SKIPIF1<0【解析】由SKIPIF1<0可得SKIPIF1<0，由SKIPIF1<0可得SKIPIF1<0，所以該直線恒過的定點(diǎn)SKIPIF1<0.故答案為：SKIPIF1<0.4．（2022·重慶）直線SKIPIF1<0經(jīng)過的定點(diǎn)坐標(biāo)是______．【答案】SKIPIF1<0【解析】把直線SKIPIF1<0的方程改寫成：SKIPIF1<0，由方程組SKIPIF1<0，解得：SKIPIF1<0，所以直線SKIPIF1<0總過定點(diǎn)SKIPIF1<0，故答案為：SKIPIF1<05．（2022江西）已知直線SKIPIF1<0為實(shí)數(shù)）過定點(diǎn)SKIPIF1<0，則點(diǎn)SKIPIF1<0的坐標(biāo)為____．【答案】SKIPIF1<0【解析】直線SKIPIF1<0為實(shí)數(shù)），即SKIPIF1<0，則SKIPIF1<0，解得SKIPIF1<0，所以直線恒過定點(diǎn)SKIPIF1<0，故答案為：SKIPIF1<0．題組五題組五三種距離1．（2022大興）直線SKIPIF1<0與直線SKIPIF1<0間的距離等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【解析】直線SKIPIF1<0即為SKIPIF1<0，直線SKIPIF1<0即為SKIPIF1<0，因?yàn)閮芍本€平行，所以距離SKIPIF1<0，故答案為：B.2．（2022朝陽）點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離是（）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】B【解析】由點(diǎn)到直線距離公式得SKIPIF1<0。故答案為：B3．（2022滕州）兩平行直線SKIPIF1<0與SKIPIF1<0之間的距離為（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】化簡直線SKIPIF1<0可得：SKIPIF1<0，根據(jù)平行線間距離公式知SKIPIF1<0。故答案為：B.4．（2022湖南）已知直線SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離SKIPIF1<0（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】由SKIPIF1<0可得SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0故答案為：D5（2022河北）已知SKIPIF1<0兩點(diǎn)到直線SKIPIF1<0的距離相等，則SKIPIF1<0（）A．2 B．SKIPIF1<0 C．2或SKIPIF1<0 D．2或SKIPIF1<0【答案】D【解析】因?yàn)镾KIPIF1<0兩點(diǎn)到直線SKIPIF1<0的距離相等，所以有SKIPIF1<0，或SKIPIF1<0，故答案為：D6．（2022浙江）已知點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，則SKIPIF1<0的最小值為．【答案】2【解析】SKIPIF1<0可以理解為點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離，又∵點(diǎn)SKIPIF1<0在直線SKIPIF1<0上，∴SKIPIF1<0的最小值等于點(diǎn)SKIPIF1<0到直線SKIPIF1<0的距離，且SKIPIF1<0.故答案為：2.7（2022內(nèi)蒙古）已知直線SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0，此時(shí)SKIPIF1<0與SKIPIF1<0之間的距離為．【答案】-1；SKIPIF1<0【解析】直線SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0．若SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0，此時(shí)SKIPIF1<0與SKIPIF1<0重合，故舍去；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0，此時(shí)SKIPIF1<0與SKIPIF1<0平行；故SKIPIF1<0；若SKIPIF1<0，即SKIPIF1<0：SKIPIF1<0，即SKIPIF1<0：SKIPIF1<0，SKIPIF1<0：SKIPIF1<0，所以SKIPIF1<0與SKIPIF1<0之間的距離為SKIPIF1<0.故答案為：-1，SKIPIF1<0.題組六題組六對(duì)稱問題1．（2022青海）在直角坐標(biāo)系中，若SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，則SKIPIF1<0的最小值是．【答案】SKIPIF1<0【解析】由題意可知，點(diǎn)SKIPIF1<0在SKIPIF1<0軸上，點(diǎn)SKIPIF1<0關(guān)于SKIPIF1<0軸的對(duì)稱點(diǎn)為SKIPIF1<0，由對(duì)稱性可得SKIPIF1<0，所以，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，故SKIPIF1<0的最小值為SKIPIF1<0.故答案為：SKIPIF1<0.2（2022云南）有一光線從點(diǎn)A(-3，5)射到直線SKIPIF1<0：3x–4y+4=0以后，再反射到點(diǎn)B(2，15)，則這條光線的入射線的反射線所在直線的方程為.【答案】SKIPIF1<0【解析】設(shè)點(diǎn)B(2，15)關(guān)于直線1：3x-4y+4=0的對(duì)稱點(diǎn)為B'(a，b)，

則15?b2?a×34=?13×2+a2?4×15+b2+4=0，解得a=14，b=-