《2025年名校题库八年级数学上册北师大版》

7. (锦江区期末)如图,在平面直角坐标系中,已知点$A(2,0)$,$B(0,-4)$,$C(-6,6)$,$D(6,6)$,以$CD为边在CD下方作正方形CDEF$.
(1)求直线$AB$的解析式;
(2)$N$为正方形边上一点,若$S_{\triangle ABN}= 8$,求点$N$的坐标;
(3)$N$为正方形边上一点,$M(0,m)为y$轴上一点,若点$N绕点M按顺时针方向旋转90^{\circ}后落在线段AB$上,请直接写出$m$的取值范围.

答案：
解：
(1)设直线$AB$的解析式为$y = kx + b$，将点$A(2,0)$，$B(0,-4)$代入，$\begin{cases}2k + b = 0,\\b = -4,\end{cases}$$\therefore\begin{cases}k = 2,\\b = -4,\end{cases}$$\therefore y = 2x - 4$。
(2)$\because C(-6,6)$，$D(6,6)$，$\therefore E(6,-6)$，$F(-6,-6)$。如图1，当点$N(6,0)$时，$AN = 4$，$OB = 4$，$\therefore S_{\triangle ABN}=\frac{1}{2}×4×4 = 8$，$\therefore N(6,0)$。如图2，过点$(6,0)$作$AB$的平行线交$EF$于点$N$。由同底等高的三角形面积相等，可得$S_{\triangle ABN}=8$。设所作平行线的解析式为$y = 2x + c$。$\because$点$(6,0)$在该直线上，$\therefore 0 = 12 + c$，$\therefore c = -12$，$\therefore y = 2x - 12$。$\because$点$N$的纵坐标为$-6$，$\therefore$点$N$的横坐标为$3$，$\therefore N(3,-6)$。如图3，当点$N$在$CD$边上时，记$BN$与$x$轴的交点为$K$，$\therefore S_{\triangle ABN}=\frac{1}{2}AK×(4 + 6)=5AK = 8$，$\therefore AK = \frac{8}{5}$，$\therefore K(\frac{2}{5},0)$。$\because$点$B(0,-4)$，易得直线$BK$的解析式为$y = 10x - 4$。$\because$点$N$的纵坐标为$6$，易得点$N$的横坐标为$1$，$\therefore N(1,6)$。如图4，当点$N$在$EF$上且在$y$轴左侧时，过点$A$作$AG\perp EF$于点$G$。设点$N(t,-6)$，则$NG = -t + 2$，$\therefore S_{\triangle ANG}=S_{\triangle ABN}+S_{\triangle NBT}+S_{梯形AGTB}$，即$\frac{1}{2}×(-t + 2)×6 = 8+\frac{1}{2}×2×(-t)+\frac{1}{2}×(2 + 6)×2$，$\therefore t = -5$，$\therefore N(-5,-6)$。综上所述，点$N$的坐标为$(6,0)$或$(3,-6)$或$(-5,-6)$或$(1,6)$。

(3)$-\frac{22}{3}\leqslant m\leqslant -6$或$2\leqslant m\leqslant\frac{14}{3}$