福建專版2024年中考數(shù)學(xué)復(fù)習(xí)第三單元函數(shù)及其圖象課時(shí)訓(xùn)練10平面直角坐標(biāo)系

PAGEPAGE1課時(shí)訓(xùn)練(十)平面直角坐標(biāo)系 (限時(shí):35分鐘)|夯實(shí)基礎(chǔ)|1.點(diǎn)P(4,3)所在的象限是 ()A.第一象限 B.其次象限C.第三象限 D.第四象限2.已知點(diǎn)P(x+3,x-4)在x軸上,則x的值為 ()A.3 B.-3 C.-4 D.43.[2024·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.已知△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),則點(diǎn)B的對(duì)應(yīng)點(diǎn)B1的坐標(biāo)為 ()A.(7,1) B.(1,7)C.(1,1) D.(2,1)6.[2024·嘉興]如圖K10-1,在直角坐標(biāo)系中,已知菱形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OA'B'C',再作圖形OA'B'C'關(guān)于點(diǎn)O的中心對(duì)稱圖形OA″B″C″,則點(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.已知點(diǎn)A(1,0),B(0,2),點(diǎn)P在x軸上,且△PAB的面積為5,則點(diǎn)P的坐標(biāo)為 ()A.(-4,0) B.(6,0)C.(-4,0)或(6,0) D.無法確定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.已知線段MN平行于y軸,且MN的長(zhǎng)度為5,若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,若菱形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,則點(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,則點(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,BC,推斷所得圖形的形態(tài)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)干脆寫出點(diǎn)B的坐標(biāo);(3)求四邊形AOCD的面積.圖K10-5|實(shí)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,則點(diǎn)P2的坐標(biāo)是 ()A.(-2,3) B.(-3,2)C.(2,-3) D.(3,-2)15.[2024·濱州]已知點(diǎn)P(a-3,2-a)關(guān)于原點(diǎn)對(duì)稱的點(diǎn)在第四象限,則a的取值范圍在數(shù)軸上表示正確的是 ()圖K10-616.如圖K10-7,點(diǎn)A,B的坐標(biāo)分別為(1,0),(0,2),若將線段AB平移至A1B1,點(diǎn)A1,B1的坐標(biāo)分別為(2,a),(b,3),則a+b=.

圖K10-717.如圖K10-8,在平面直角坐標(biāo)系中,已知點(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,則P點(diǎn)的坐標(biāo)是.

圖K10-8|思維拓展|18.[2024·綿陽]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°,則對(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.[2024·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,圓心角為120°的AB多次復(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))動(dòng)身,以每秒23π米的速度沿曲線向右運(yùn)動(dòng),則在第2024秒時(shí)點(diǎn)P的縱坐標(biāo)為 (圖K10-10A.-2 B.-1 C.0 D.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),已知點(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)若點(diǎn)A1的坐標(biāo)為(2,1),則點(diǎn)A3的坐標(biāo)為,點(diǎn)A2024的坐標(biāo)為;

(2)若A2024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),若點(diǎn)A1,A2,A3,…,An均在y軸左側(cè),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)如圖所示,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=S△AOD+S△ODC=12×2×214.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)在其次象限,∴a-3<0,2-a>016.217.(-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-解得k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]過點(diǎn)E作EF⊥x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=12×4∴AF=12AE=∴OF=AO-AF=4-1=3,EF=3,故選D.19.B[解析]半徑為2米,圓心角為120°的弧長(zhǎng)為:120×π×2180=4π點(diǎn)P從原點(diǎn)A動(dòng)身,沿這條曲線向右運(yùn)動(dòng),速度為每秒23當(dāng)點(diǎn)P從原點(diǎn)O動(dòng)身,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),…,依據(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)P2024的坐標(biāo).∵2024÷4=504…3,∴P2024的坐標(biāo)是(20243,-1),∴在第2024秒時(shí)點(diǎn)P的縱坐標(biāo)為-1.故答案為B.20.解:(1)視察發(fā)覺規(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).∵2024=505×4,∴點(diǎn)A2024的坐標(biāo)為(-2,3