《2025年名校题库八年级数学上册北师大版》

6. (武侯区期末)在平面直角坐标系$xOy$中,已知点$M(-2,-2)$,过点$M作直线AB$,交$x轴的负半轴于点A$,交$y轴的负半轴于点B(0,m)$.
(1)如图1,当$m= -6$时.
①求直线$AB$的函数表达式;
②过点$A作y轴的平行线l$,$N是l$上一动点,连接$BN$,$MN$,若$S_{\triangle MBN}= \frac{3}{8}S_{\triangle ABO}$,求满足条件的点$N$的坐标.
(2)如图2,将直线$AB绕点B顺时针旋转45^{\circ}$后,交$x轴的正半轴于点C$,过点$C作CD\perp BC$,交直线$AB于点D$.试问:随着$m$值的改变,点$D$的横坐标是否发生变化?若不变,求出点$D$的横坐标;若变化,请说明理由.

答案：
解：
(1)①$\because m = -6$，$\therefore B(0,-6)$，$\therefore$设直线$AB$的函数表达式为$y = kx - 6$。$\because$点$M(-2,-2)$在直线$AB$上，$-2 = -2k - 6$，$\therefore k = -2$，$\therefore$直线$AB$的函数表达式为$y = -2x - 6$。②如图1，由①知，直线$AB$的函数表达式为$y = -2x - 6$，令$y = 0$，则$-2x - 6 = 0$，$\therefore x = -3$，$\therefore A(-3,0)$，$\therefore$直线$l$为$x = -3$，$\therefore$设$N(-3,t)$，$\therefore AN = |t|$。$\because A(-3,0)$，$B(0,-6)$，$\therefore OA = 3$，$OB = 6$，$\therefore S_{\triangle ABO}=\frac{1}{2}OA\cdot OB=\frac{1}{2}×3×6 = 9$。$\because S_{\triangle MBN}=\frac{3}{8}S_{\triangle ABO}$，$\therefore S_{\triangle MBN}=\frac{3}{8}×9=\frac{27}{8}$。过点$M$作$MF\perp AN$于点$F$，过点$B$作$BE\perp AN$于点$E$，$\therefore MF = 1$，$BE = 3$，$\therefore S_{\triangle MBN}=S_{\triangle BAN}-S_{\triangle AMN}=\frac{1}{2}AN\cdot BE-\frac{1}{2}AN\cdot FM=\frac{1}{2}|t|(3 - 1)=|t|=\frac{27}{8}$，$\therefore t = \pm\frac{27}{8}$，$\therefore N(-3,\frac{27}{8})$或$(-3,-\frac{27}{8})$。

(2)如图2，$\because\angle ABC = 45^{\circ}$，$\angle BCD = 90^{\circ}$，$\therefore\angle ADC = 45^{\circ}=\angle ABC$，$\therefore CD = CB$，$\therefore\triangle BDC$是等腰直角三角形。$\because M(-2,-2)$，$B(0,m)$，$\therefore$直线$AB$的函数表达式为$y = \frac{m + 2}{2}x + m$。设点$C(a,0)$，分别过点$D$，$B$作$y$轴的垂线，过点$C$作$x$轴的垂线，交前两条直线于点$G$，$H$，则$\angle H = \angle G = \angle OCH = \angle OBH = 90^{\circ}$，$\therefore$四边形$OBHC$为矩形，$\therefore OC = BH$。$\because\angle G = \angle BCD = 90^{\circ}$，$\therefore\angle CDG+\angle DCG=\angle DCG+\angle BCH = 90^{\circ}$，$\therefore\angle CDG = \angle BCH$，$\therefore\triangle DCG\cong\triangle CBH(AAS)$，$\therefore BH = OC = CG = a$，$CH = DG = -m$，$\therefore D(m + a,a)$，$\therefore a = \frac{m + 2}{2}\cdot(m + a)+m$，$\therefore m^{2}+ma + 4m = 0$。$\because m\neq0$，$\therefore m + a = -4$，即点$D$的横坐标为$-4$，保持不变。