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北師大八級(jí)上第章位置與坐標(biāo)單元測(cè)試含答案解析(參考版)

2025-01-17 17:07本頁(yè)面
  

【正文】 后的圖形..則C′(2,1),即旋轉(zhuǎn)后點(diǎn)C的坐標(biāo)是(2,1).故答案是:(2,1).【點(diǎn)評(píng)】本題考查了利用旋轉(zhuǎn)變換作圖,熟練掌握網(wǎng)格結(jié)構(gòu)準(zhǔn)確找出對(duì)應(yīng)點(diǎn)的位置是解題的關(guān)鍵. 22.設(shè)點(diǎn)M(1,2)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)為M′,則M′的坐標(biāo)為?。ī?,﹣2)?。究键c(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo).【分析】根據(jù)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)特點(diǎn):兩個(gè)點(diǎn)關(guān)于原點(diǎn)對(duì)稱時(shí),它們的坐標(biāo)符號(hào)相反可直接得到答案.【解答】解:點(diǎn)M(1,2)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)M′的坐標(biāo)為(﹣1,﹣2),故答案為:(﹣1,﹣2).【點(diǎn)評(píng)】此題主要考查了關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)特點(diǎn),關(guān)鍵是熟練掌握點(diǎn)的坐標(biāo)的變化規(guī)律. 23.已知點(diǎn)M(3,﹣2),將它先向左平移4個(gè)單位,再向上平移3個(gè)單位后得到點(diǎn)N,則點(diǎn)N的坐標(biāo)是?。ī?,1) .【考點(diǎn)】坐標(biāo)與圖形變化平移.【分析】直接利用平移中點(diǎn)的變化規(guī)律求解即可.平移中點(diǎn)的變化規(guī)律是:橫坐標(biāo)右移加,左移減;縱坐標(biāo)上移加,下移減.【解答】解:原來(lái)點(diǎn)的橫坐標(biāo)是3,縱坐標(biāo)是﹣2,向左平移4個(gè)單位,再向上平移3個(gè)單位得到新點(diǎn)的橫坐標(biāo)是3﹣4=﹣1,縱坐標(biāo)為﹣2+3=1.則點(diǎn)N的坐標(biāo)是(﹣1,1).故答案填:(﹣1,1).【點(diǎn)評(píng)】解題關(guān)鍵是要懂得左右平移點(diǎn)的縱坐標(biāo)不變,而上下平移時(shí)點(diǎn)的橫坐標(biāo)不變,平移變換是中考的??键c(diǎn),平移中點(diǎn)的變化規(guī)律是:橫坐標(biāo)右移加,左移減;縱坐標(biāo)上移加,下移減. 24.點(diǎn)P(5,﹣3)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)的坐標(biāo)為?。ī?,3)?。究键c(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo).【分析】?jī)牲c(diǎn)關(guān)于原點(diǎn)對(duì)稱,橫坐標(biāo)互為相反數(shù),縱坐標(biāo)互為相反數(shù).【解答】解:∵5的相反數(shù)是﹣5,﹣3的相反數(shù)是3,∴點(diǎn)P(5,﹣3)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)的坐標(biāo)為 (﹣5,3),故答案為:(﹣5,3).【點(diǎn)評(píng)】主要考查兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱的坐標(biāo)的特點(diǎn):兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱,兩點(diǎn)的橫坐標(biāo)互為相反數(shù),縱坐標(biāo)互為相反數(shù),用到的知識(shí)點(diǎn)為:a的相反數(shù)為﹣a. 25.在平面直角坐標(biāo)系中,點(diǎn)(﹣3,2)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)是?。?,﹣2) .【考點(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo).【專(zhuān)題】數(shù)形結(jié)合.【分析】根據(jù)平面直角坐標(biāo)系內(nèi)兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱橫縱坐標(biāo)互為相反數(shù),即可得出答案.【解答】解:根據(jù)平面直角坐標(biāo)系內(nèi)兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱橫縱坐標(biāo)互為相反數(shù),∴點(diǎn)(﹣3,2)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)是(3,﹣2),故答案為(3,﹣2).【點(diǎn)評(píng)】本題主要考查了平面直角坐標(biāo)系內(nèi)兩點(diǎn)關(guān)于原點(diǎn)對(duì)稱橫縱坐標(biāo)互為相反數(shù),難度較小. 26.已知點(diǎn)P(3,2),則點(diǎn)P關(guān)于y軸的對(duì)稱點(diǎn)P1的坐標(biāo)是?。ī?,2) ,點(diǎn)P關(guān)于原點(diǎn)O的對(duì)稱點(diǎn)P2的坐標(biāo)是?。ī?,﹣2)?。究键c(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo);關(guān)于x軸、y軸對(duì)稱的點(diǎn)的坐標(biāo).【分析】根據(jù)關(guān)于y軸對(duì)稱的點(diǎn)的橫坐標(biāo)互為相反數(shù),縱坐標(biāo)相同;關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的橫坐標(biāo)與縱坐標(biāo)都互為相反數(shù)解答.【解答】解:點(diǎn)P(3,2)關(guān)于y軸的對(duì)稱點(diǎn)P1的坐標(biāo)是(﹣3,2),點(diǎn)P關(guān)于原點(diǎn)O的對(duì)稱點(diǎn)P2的坐標(biāo)是(﹣3,﹣2).故答案為:(﹣3,2);(﹣3,﹣2).【點(diǎn)評(píng)】本題考查了關(guān)于原點(diǎn)對(duì)稱點(diǎn)點(diǎn)的坐標(biāo),關(guān)于y軸對(duì)稱的點(diǎn)的坐標(biāo),熟記對(duì)稱點(diǎn)的坐標(biāo)特征是解題的關(guān)鍵. 27.在平面直角坐標(biāo)系中,點(diǎn)P(5,﹣3)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)是 (﹣5,3) .【考點(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo).【分析】根據(jù)關(guān)于坐標(biāo)原點(diǎn)對(duì)稱的點(diǎn)的橫坐標(biāo)與縱坐標(biāo)都互為相反數(shù)解答.【解答】解:點(diǎn)P(5,﹣3)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)是(﹣5,3).故答案為:(﹣5,3).【點(diǎn)評(píng)】本題考查了關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo),熟記關(guān)于坐標(biāo)原點(diǎn)對(duì)稱的點(diǎn)的橫坐標(biāo)與縱坐標(biāo)都互為相反數(shù)是解題的關(guān)鍵. 28.若將等腰直角三角形AOB按如圖所示放置,OB=2,則點(diǎn)A關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)為?。ī?,﹣1) .【考點(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo).【分析】過(guò)點(diǎn)A作AD⊥OB于點(diǎn)D,根據(jù)等腰直角三角形的性質(zhì)求出OD及AD的長(zhǎng),故可得出A點(diǎn)坐標(biāo),再由關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)特點(diǎn)即可得出結(jié)論.【解答】解:過(guò)點(diǎn)A作AD⊥OB于點(diǎn)D,∵△AOB是等腰直角三角形,OB=2,∴OD=AD=1,∴A(1,1),∴點(diǎn)A關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)為(﹣1,﹣1).故答案為(﹣1,﹣1).【點(diǎn)評(píng)】本題考查的是關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo)特點(diǎn),熟知等腰直角三角形的性質(zhì)是解答此題的關(guān)鍵. 三、解答題(共2小題)29.在平面直角坐標(biāo)系xOy中,已知A(﹣1,5),B(4,2),C(﹣1,0)三點(diǎn).(1)點(diǎn)A關(guān)于原點(diǎn)O的對(duì)稱點(diǎn)A′的坐標(biāo)為?。?,﹣5) ,點(diǎn)B關(guān)于x軸的對(duì)稱點(diǎn)B′的坐標(biāo)為 (4,﹣2) ,點(diǎn)C關(guān)于y軸的對(duì)稱點(diǎn)C的坐標(biāo)為?。?,0)?。?)求(1)中的△A′B′C′的面積.【考點(diǎn)】關(guān)于原點(diǎn)對(duì)稱的點(diǎn)的坐標(biāo);三角形的面積;關(guān)于x軸、y軸對(duì)稱的點(diǎn)的坐標(biāo).【分析】(1)關(guān)于原點(diǎn)對(duì)稱的兩點(diǎn)的橫、縱坐標(biāo)都是互為相反數(shù);關(guān)于x軸對(duì)稱的兩點(diǎn)的橫坐標(biāo)相同,縱坐標(biāo)互為相反數(shù);關(guān)于y軸對(duì)稱的兩點(diǎn)的橫坐標(biāo)互為相反數(shù),縱坐標(biāo)相同;(2)根據(jù)點(diǎn)A′(1,﹣5),B′(4,﹣2),C′(1,0)在平面直角坐標(biāo)系中的位置,可以求得A′C′=5,B′D=3,所以由三角形的面積公式進(jìn)行解答.【解答】解:(1)∵A(﹣1,5),∴點(diǎn)A關(guān)于原點(diǎn)O的對(duì)稱點(diǎn)A′的坐標(biāo)為(1,﹣5).∵B(4,2),∴點(diǎn)B關(guān)于x軸的對(duì)稱點(diǎn)B′的坐標(biāo)為(4,﹣2).∵C(﹣1,0),∴點(diǎn)C關(guān)于y軸的對(duì)稱點(diǎn)C′的坐標(biāo)為(1,0).故答案為:(1,﹣5),(4,﹣2),(1,0).(2)如圖,∵A′(1,﹣5),B′(4,﹣2),C′(1,0).∴A′C′=|﹣5﹣0|=5,B′D=|4﹣1|=3,∴S△A′B′C′=A′C′?B′D=53=,即(1)中的△A′B′C′.【點(diǎn)評(píng)】本題考查了關(guān)于原點(diǎn)、x軸、y
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