《2025年名校题库八年级数学上册北师大版》

1. (武侯区期末)如图,在平面直角坐标系中,一次函数$y= 2x+4的图象分别与x$轴、$y轴相交于A$,$B$两点.将直线$AB绕点A逆时针旋转45^{\circ}$后,与$y轴交于点C$,则点$C$的坐标为____.

答案：
$(0,-6)$ [解析]由一次函数$y = 2x + 4$的图象分别与$x$轴、$y$轴相交于$A$，$B$两点，易得$A(-2,0)$，$B(0,4)$，$\therefore OA = 2$，$OB = 4$。如图，作$DB\perp AB$交直线$AC$于点$D$，过点$D$作$DE\perp y$轴于点$E$。
$\because\angle BAD = 45^{\circ}$，$\therefore\triangle BAD$是等腰直角三角形，$\therefore AB = DB$。$\because\angle OAB+\angle ABO=\angle ABO+\angle DBE = 90^{\circ}$，$\therefore\angle OAB=\angle DBE$。在$\triangle ABO$和$\triangle BDE$中，$\begin{cases}\angle OAB = \angle DBE,\\\angle AOB = \angle BED,\\AB = DB,\end{cases}$$\therefore\triangle ABO\cong\triangle BDE(AAS)$，$\therefore BE = OA = 2$，$DE = BO = 4$，$\therefore D(-4,6)$。设直线$AC$的函数表达式为$y = kx + b$，把点$A$，$D$的坐标代入，得$\begin{cases}-2k + b = 0,\\-4k + b = 6,\end{cases}$解得$\begin{cases}k = -3,\\b = -6,\end{cases}$$\therefore$直线$AC$的函数表达式为$y = -3x - 6$，$\therefore$点$C$的坐标为$(0,-6)$。

2. (金牛区期末)如图,在平面直角坐标系中,已知直线$y= x上一点P(2,2)$,$C为y$轴上一点,连接$PC$,线段$PC绕点P顺时针旋转90^{\circ}至线段PD$,过点$D作直线AB\perp x$轴,垂足为$B$,直线$AB与直线y= x交于点A$,连接$CD$,直线$CD与直线y= x交于点Q$.当$\triangle OPC\cong \triangle ADP$时,点$C$的坐标是____,点$Q$的坐标是____.

答案：
$(0,4 + 2\sqrt{2})$ $(2\sqrt{2}+2,2\sqrt{2}+2)$ [解析]如图，过点$P$作$x$轴的平行线交$y$轴于点$M$，交$AB$于点$N$。设$C(0,t)$。$\because P(2,2)$，$\therefore OP = 2\sqrt{2}$，$OM = BN = PM = 2$，$CM = t - 2$。$\because$线段$PC$绕点$P$顺时针旋转$90^{\circ}$至线段$PD$，$\therefore PC = PD$，$\angle CPD = 90^{\circ}$，$\therefore\angle CPM+\angle DPN = 90^{\circ}$，而$\angle CPM+\angle PCM = 90^{\circ}$，$\therefore\angle PCM=\angle DPN$。在$\triangle PCM$和$\triangle DPN$中，$\begin{cases}\angle PMC = \angle DNP,\\\angle PCM = \angle DPN,\\PC = DP,\end{cases}$$\therefore\triangle PCM\cong\triangle DPN$，$\therefore PN = CM = t - 2$，$DN = PM = 2$，$\therefore MN = t - 2 + 2 = t$，$DB = 2 + 2 = 4$，$\therefore D(t,4)$。$\because\triangle OPC\cong\triangle ADP$，$\therefore AD = OP = 2\sqrt{2}$，$\therefore A(t,4 + 2\sqrt{2})$。把$A(t,4 + 2\sqrt{2})$代入$y = x$，得$t = 4 + 2\sqrt{2}$，$\therefore C(0,4 + 2\sqrt{2})$，$D(4 + 2\sqrt{2},4)$。设直线$CD$的解析式为$y = kx + b$，把$C(0,4 + 2\sqrt{2})$，$D(4 + 2\sqrt{2},4)$代入，得$\begin{cases}b = 4 + 2\sqrt{2},\\(4 + 2\sqrt{2})k + b = 4,\end{cases}$解得$\begin{cases}k = 1-\sqrt{2},\\b = 4 + 2\sqrt{2},\end{cases}$$\therefore$直线$CD$的解析式为$y = (1-\sqrt{2})x + 4 + 2\sqrt{2}$。联立方程组$\begin{cases}y = x,\\y = (1-\sqrt{2})x + 4 + 2\sqrt{2},\end{cases}$解得$\begin{cases}x = 2\sqrt{2}+2,\\y = 2\sqrt{2}+2,\end{cases}$$\therefore Q(2\sqrt{2}+2,2\sqrt{2}+2)$。

3. (石室联中)如图,在平面直角坐标系中,直线$y= -\frac{\sqrt{3}}{3}x+4与x$轴、$y轴分别交于点M$,$N$,点$A$,$B分别在y$轴、$x$轴上,且$\angle ABO= 30^{\circ}$,$AB= 4$.将$\triangle ABO绕原点O$顺时针转动一周,当$AB与直线MN$平行时,点$A$的坐标为____.

答案：
$(\sqrt{3},1)$或$(-\sqrt{3},-1)$ [解析]①如图1，过点$A$作$AC\perp x$轴于点$C$，$\because AB = 4$，$\angle ABO = 30^{\circ}$，$\therefore OA = \frac{1}{2}AB = 2$，$\angle BAO = 90^{\circ}-30^{\circ}=60^{\circ}$，$\therefore\angle OAD = 120^{\circ}$。$\because$直线$MN$的解析式为$y = -\frac{\sqrt{3}}{3}x + 4$，$\therefore\angle NMO = 30^{\circ}$。$\because AB// MN$，$\therefore\angle ADO = \angle NMD = 30^{\circ}$，$\therefore\angle AOC = 30^{\circ}$，$\therefore AC = \frac{1}{2}OA = 1$，$\therefore OC = \sqrt{2^{2}-1^{2}}=\sqrt{3}$，$\therefore$点$A$的坐标为$(\sqrt{3},1)$。②如图2，易得图2中的点$A$与图1中的点$A$关于原点对称，$\therefore$此时点$A$的坐标为$(-\sqrt{3},-1)$。综上所述，点$A$的坐标为$(\sqrt{3},1)$或$(-\sqrt{3},-1)$。

4. (青羊区期末)如图,在平面直角坐标系中,直线$y= -\sqrt{3}x+4与x$轴、$y轴分别交于点M$,$N$,点$A$,$B分别在y$轴、$x$轴上,且$\angle BAO= 30^{\circ}$,$AO= 2$.将$\triangle ABO绕原点O$顺时针转动一周,当$AB与直线MN$垂直时,点$A$的坐标为____.

答案：
$(1,\sqrt{3})$或$(-1,-\sqrt{3})$ [解析]当$x = 0$时，$y = -\sqrt{3}x + 4 = 4$，则$N(0,4)$。当$y = 0$时，$-\sqrt{3}x + 4 = 0$，解得$x = \frac{4\sqrt{3}}{3}$，则$M(\frac{4\sqrt{3}}{3},0)$。在$Rt\triangle OMN$中，$\because ON = 4$，$OM = \frac{4\sqrt{3}}{3}$，$\therefore MN = \sqrt{ON^{2}+OM^{2}}=\frac{8\sqrt{3}}{3}$，$\therefore OM = \frac{1}{2}MN$，$\therefore\angle MNO = 30^{\circ}$，$\therefore\angle NMO = 60^{\circ}$。在$Rt\triangle ABO$中，$\because\angle BAO = 30^{\circ}$，$AO = 2$，$\therefore\angle OBA = 60^{\circ}$，$OB = \frac{2\sqrt{3}}{3}$。$\because AB$与直线$MN$垂直，$\therefore$直线$AB$与$x$轴的夹角为$30^{\circ}$。①如图1，设直线$AB$交$y$轴于点$C$，交$MN$于点$G$，过点$A$作$AD\perp x$轴于点$D$，过点$G$作$GH\perp x$轴于点$H$，$\therefore\angle MGH = 30^{\circ}$，$\therefore\angle BGH = 60^{\circ}$，$\therefore\angle OCB = 60^{\circ}$。又$\because\angle OBA = 60^{\circ}$，$\therefore\triangle OBC$是等边三角形，$\therefore\angle BOC = 60^{\circ}$，$\therefore\angle AOC = 30^{\circ}$，$\therefore\angle AOD = 60^{\circ}$，$\therefore$在$Rt\triangle OAD$中，$OD = \frac{1}{2}OA = 1$，$AD = \frac{\sqrt{3}}{2}OA = \sqrt{3}$，$\therefore$此时点$A$的坐标为$(1,\sqrt{3})$。②如图2，设直线$AB$交$y$轴于点$C$，过点$A$作$AD\perp x$轴于点$D$，与①同理可得$\angle OCB = 60^{\circ}$。又$\because\angle ABO = 60^{\circ}$，$\therefore\angle COB = 60^{\circ}$，$\therefore\angle AOC = 30^{\circ}$，$\therefore\angle AOD = 60^{\circ}$，$\therefore$在$Rt\triangle OAD$中，$OD = \frac{1}{2}OA = 1$，$AD = \frac{\sqrt{3}}{2}OA = \sqrt{3}$，$\therefore$此时点$A$的坐标为$(-1,-\sqrt{3})$。综上所述，点$A$的坐标为$(1,\sqrt{3})$或$(-1,-\sqrt{3})$。

5. (新都区期末)如图1,在平面直角坐标系中,点$A的坐标为(4,4)$,点$B的坐标为(-4,0)$.
(1)求直线$AB$的表达式;
(2)$M$是坐标轴上的一点,若以$AB为直角边构造Rt\triangle ABM$,请求出满足条件的所有点$M$的坐标;
(3)如图2,以$A为直角顶点作\angle CAD= 90^{\circ}$,射线$AC交x轴的正半轴于点C$,射线$AD交y轴的负半轴于点D$,当$\angle CAD绕点A$旋转时,求$OC-OD$的值.

答案：
解：
(1)设直线$AB$的表达式为$y = kx + b(k\neq0)$。$\because$点$A(4,4)$，$B(-4,0)$在直线$AB$上，$\begin{cases}4k + b = 4,\\-4k + b = 0,\end{cases}$解得$\begin{cases}k = \frac{1}{2},\\b = 2,\end{cases}$$\therefore$直线$AB$的表达式为$y = \frac{1}{2}x + 2$。
(2)$\because\triangle ABM$是以$AB$为直角边的直角三角形，$\therefore$有$\angle BAM = 90^{\circ}$或$\angle ABM = 90^{\circ}$。
①当$\angle BAM = 90^{\circ}$时，如图1，过点$A$作$AB$的垂线，交$y$轴于点$M_{1}$，交$x$轴于点$M_{2}$，设$M_{1}M_{2}$所在直线的表达式为$y = ax + c$。$\because M_{1}M_{2}$所在直线与$AB$垂直，$\therefore a = -2$。$\because$直线$M_{1}M_{2}$过点$A(4,4)$，代入得$-2×4 + c = 4$，解得$c = 12$，$\therefore y = -2x + 12$。令$y = 0$，得$-2x + 12 = 0$，解得$x = 6$，$\therefore M_{2}(6,0)$。令$x = 0$，得$y = 12$，$\therefore M_{1}(0,12)$。

②当$\angle ABM = 90^{\circ}$时，如图2，过点$B$作$AB$的垂线交$y$轴于点$M_{3}$，同理可得$BM_{3}$所在直线的表达式为$y = -2x - 8$，令$x = 0$，得$y = -8$，$\therefore M_{3}(0,-8)$。综上所述，点$M$的坐标为$(6,0)$或$(0,12)$或$(0,-8)$。
(3)如图3，过点$A$分别作$x$轴、$y$轴的垂线，垂足分别为$G$，$H$，则$\angle AGC = \angle AHD = 90^{\circ}$。又$\because\angle HOC = 90^{\circ}$，$\therefore\angle GAH = 90^{\circ}$，$\therefore\angle DAG+\angle DAH = 90^{\circ}$。$\because\angle CAD = 90^{\circ}$，$\therefore\angle DAG+\angle CAG = 90^{\circ}$，$\therefore\angle CAG = \angle DAH$。$\because A(4,4)$，$\therefore OG = AH = AG = OH = 4$。在$\triangle AGC$和$\triangle AHD$中，$\begin{cases}\angle AGC = \angle AHD,\\AG = AH,\\\angle CAG = \angle DAH,\end{cases}$$\therefore\triangle AGC\cong\triangle AHD(ASA)$，$\therefore GC = HD$，$\therefore OC - OD=(OG + GC)-(HD - OH)=OG + OH = 8$。