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1、課時(shí)訓(xùn)練(十)平面直角坐標(biāo)系(限時(shí):35分鐘)|夯實(shí)根底|1.點(diǎn)P(4,3)所在的象限是()A.第一象限B.第二象限C.第三象限D(zhuǎn).第四象限2.點(diǎn)P(x+3,x-4)在x軸上,那么x的值為()A.3B.-3C.-4D.43.2022常德點(diǎn)(-1,2)關(guān)于原點(diǎn)的對(duì)稱點(diǎn)坐標(biāo)是()A.(-1,-2)B.(1,-2)C.(1,2)D.(2,-1)4.平面直角坐標(biāo)系內(nèi)的點(diǎn)A(-1,2)與點(diǎn)B(-1,-2)關(guān)于()A.y軸對(duì)稱B.x軸對(duì)稱C.原點(diǎn)對(duì)稱D.直線y=x對(duì)稱5.ABC的頂點(diǎn)坐標(biāo)分別是A(0,6),B(-3,-3),C(1,0),將ABC平移后頂點(diǎn)A的對(duì)應(yīng)點(diǎn)A1的坐標(biāo)是(4,10),那么點(diǎn)B的對(duì)應(yīng)

2、點(diǎn)B1的坐標(biāo)為()A.(7,1)B.(1,7)C.(1,1)D.(2,1)6.2022嘉興如圖K10-1,在直角坐標(biāo)系中,菱形OABC的頂點(diǎn)A(1,2),B(3,3).作菱形OABC關(guān)于y軸的對(duì)稱圖形OABC,再作圖形OABC關(guān)于點(diǎn)O的中心對(duì)稱圖形OABC,那么點(diǎn)C的對(duì)應(yīng)點(diǎn)C的坐標(biāo)是()圖K10-1A.(2,-1)B.(1,-2)C.(-2,1)D.(-2,-1)7.點(diǎn)A(1,0),B(0,2),點(diǎn)P在x軸上,且PAB的面積為5,那么點(diǎn)P的坐標(biāo)為()A.(-4,0)B.(6,0)C.(-4,0)或(6,0)D.無(wú)法確定8.在第四象限到x軸距離為5,到y(tǒng)軸距離為2的點(diǎn)的坐標(biāo)是.9.線段MN平行于

3、y軸,且MN的長(zhǎng)度為5,假設(shè)M(2,-2),那么點(diǎn)N的坐標(biāo)是.10.如圖K10-2,在平面直角坐標(biāo)系xOy中,假設(shè)菱形ABCD的頂點(diǎn)A,B的坐標(biāo)分別為(-3,0),(2,0),點(diǎn)D在y軸上,那么點(diǎn)C的坐標(biāo)是.圖K10-211.如圖K10-3,在平面直角坐標(biāo)系中,B,C兩點(diǎn)的坐標(biāo)分別為(-3,0)和(7,0),AB=AC=13,那么點(diǎn)A的坐標(biāo)為.圖K10-312.如圖K10-4,在平面直角坐標(biāo)系中,線段AB的兩個(gè)端點(diǎn)分別為A(-3,0),B(0,4).(1)畫出線段AB先向右平移3個(gè)單位長(zhǎng)度,再向下平移4個(gè)單位長(zhǎng)度后得到的線段CD,并寫出點(diǎn)A的對(duì)應(yīng)點(diǎn)D的坐標(biāo),點(diǎn)B的對(duì)應(yīng)點(diǎn)C的坐標(biāo);(2)連接AD

4、,BC,判斷所得圖形的形狀并求其面積.圖K10-413.如圖K10-5,四邊形OABC是矩形,且AOM=120,CO=3,BC=1.(1)求A,C兩點(diǎn)的坐標(biāo);(2)直接寫出點(diǎn)B的坐標(biāo);(3)求四邊形AOCD的面積.圖K10-5|能力提升|14.在平面直角坐標(biāo)系中,點(diǎn)P(-4,2)向右平移7個(gè)單位長(zhǎng)度得到點(diǎn)P1,點(diǎn)P1繞原點(diǎn)逆時(shí)針旋轉(zhuǎn)90得到點(diǎn)P2,那么點(diǎn)P2的坐標(biāo)是()A.(-2,3)B.(-3,2)C.(2,-3)D.(3,-2)15.2022濱州點(diǎn)P(a-3,2-a)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)在第四象限,那么a的取值范圍在數(shù)軸上表示正確的選項(xiàng)是()圖K10-616.如圖K10-7,點(diǎn)A,B的坐標(biāo)分別

5、為(1,0),(0,2),假設(shè)將線段AB平移至A1B1,點(diǎn)A1,B1的坐標(biāo)分別為(2,a),(b,3),那么a+b=.圖K10-717.如圖K10-8,在平面直角坐標(biāo)系中,點(diǎn)A(2,3),點(diǎn)B(-2,1),在x軸上存在點(diǎn)P到A,B兩點(diǎn)的距離之和最小,那么P點(diǎn)的坐標(biāo)是.圖K10-8|思維拓展|18.2022綿陽(yáng)如圖K10-9,在平面直角坐標(biāo)系中,四邊形OABC為菱形,O(0,0),A(4,0),AOC=60,那么對(duì)角線交點(diǎn)E的坐標(biāo)為()圖K10-9A.(2,3)B.(3,2)C.(3,3)D.(3,3)19.2022婁底如圖K10-10,在單位長(zhǎng)度為1米的平面直角坐標(biāo)系中,曲線是由半徑為2米,圓

6、心角為120的AB屢次復(fù)制并首尾連接而成.現(xiàn)有一點(diǎn)P從A(A為坐標(biāo)原點(diǎn))出發(fā),以每秒23米的速度沿曲線向右運(yùn)動(dòng),那么在第2022秒時(shí)點(diǎn)P的縱坐標(biāo)為()圖K10-10A.-2B.-1C.0D.120.在平面直角坐標(biāo)系xOy中,對(duì)于點(diǎn)P(x,y),我們把P(y-1,-x-1)叫做點(diǎn)P的友好點(diǎn),點(diǎn)A1的友好點(diǎn)為A2,點(diǎn)A2的友好點(diǎn)為A3,點(diǎn)A3的友好點(diǎn)為A4,這樣依次得到點(diǎn).(1)假設(shè)點(diǎn)A1的坐標(biāo)為(2,1),那么點(diǎn)A3的坐標(biāo)為,點(diǎn)A2022的坐標(biāo)為;(2)假設(shè)A2022的坐標(biāo)為(-3,2),設(shè)A1(x,y),求x+y的值;(3)設(shè)點(diǎn)A1的坐標(biāo)為(a,b),假設(shè)點(diǎn)A1,A2,A3,An均在y軸左側(cè),

7、求a,b的取值范圍.【參考答案】1.A2.D3.B4.B5.C6.A7.C8.(2,-5)9.(2,3)或(2,-7)10.(5,4)11.(2,12)12.解:(1)如下圖,D(0,-4),C(3,0).(2)四邊形ABCD是菱形,S菱形ABCD=24.13.解:(1)如圖,作兩條垂線CD,AE,垂足分別為D,E,易知A-12,32,C32,32.(2)B(1,3).(3)易知OD=233,所以S四邊形OADC=SAOD+SODC=122233=233.14.A15.C解析點(diǎn)P(a-3,2-a)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)在第四象限,點(diǎn)P(a-3,2-a)在第二象限,a-30,解得a3,a2,不等式組的

8、解集是a2,在數(shù)軸上表示如選項(xiàng)C所示.應(yīng)選C.16.217.(-1,0)解析作A關(guān)于x軸的對(duì)稱點(diǎn)C,連接BC交x軸于P,此時(shí)AP+BP最小.A點(diǎn)的坐標(biāo)為(2,3),C(2,-3),設(shè)直線BC的解析式是y=kx+b,把B,C的坐標(biāo)分別代入得-2k+b=1,2k+b=-3,解得k=-1,b=-1,即直線BC的解析式是y=-x-1,令y=0,即-x-1=0,解得x=-1,P點(diǎn)的坐標(biāo)是(-1,0).18.D解析過(guò)點(diǎn)E作EFx軸于點(diǎn)F,四邊形OABC為菱形,AOC=60,AOE=30,FAE=60,A(4,0),OA=4,AE=12AO=124=2,AF=12AE=1,OF=AO-AF=4-1=3,EF

9、=3,應(yīng)選D.19.B解析半徑為2米,圓心角為120的弧長(zhǎng)為:1202180=43(米),點(diǎn)P從原點(diǎn)A出發(fā),沿這條曲線向右運(yùn)動(dòng),速度為每秒23米,當(dāng)點(diǎn)P從原點(diǎn)O出發(fā),沿這條曲線向右運(yùn)動(dòng),運(yùn)動(dòng)時(shí)間為1秒時(shí),點(diǎn)P的坐標(biāo)為(3,1),運(yùn)動(dòng)時(shí)間為2秒時(shí),點(diǎn)P的坐標(biāo)為(23,0),運(yùn)動(dòng)時(shí)間為3秒時(shí),點(diǎn)P的坐標(biāo)為(33,-1),運(yùn)動(dòng)時(shí)間為4秒時(shí),點(diǎn)P的坐標(biāo)為(43,0),運(yùn)動(dòng)時(shí)間為5秒時(shí),點(diǎn)P的坐標(biāo)為(53,1),運(yùn)動(dòng)時(shí)間為6秒時(shí),點(diǎn)P的坐標(biāo)為(63,0),根據(jù)圖象可得移動(dòng)4秒圖象完成一個(gè)循環(huán),從而可得出點(diǎn)P2022的坐標(biāo).20224=5043,P2022的坐標(biāo)是(20223,-1),在第2022秒時(shí)點(diǎn)P

10、的縱坐標(biāo)為-1.故答案為B.20.解:(1)觀察發(fā)現(xiàn)規(guī)律:A1(2,1),A2(0,-3),A3(-4,-1),A4(-2,3),A5(2,1),A4n+1(2,1),A4n+2(0,-3),A4n+3(-4,-1),A4n+4(-2,3),n為自然數(shù).2022=5054,點(diǎn)A2022的坐標(biāo)為(-2,3).故答案為(-4,-1);(-2,3).(2)A2022的坐標(biāo)為(-3,2),A2021(1,2),A1(1,2),x+y=3.(3)A1(a,b),A2(b-1,-a-1),A3(-a-2,-b),A4(-b-1,a+1),點(diǎn)A1,A2,A3,An均在y軸左側(cè),a0,-a-20且b-10,-b-10,解得-2a0,-1b1.