《2025年名校题库八年级数学上册北师大版》

10. (高新区期末)在平面直角坐标系中,一次函数$y= -\frac {2}{3}x+4$的图象与x轴和y轴分别交于A,B两点,动点P从点A出发,在线段AO上以每秒1个单位长度的速度向点O作匀速运动,到达点O即停止运动,其中A,Q两点关于点P对称,以线段PQ为边向上作正方形PQMN.设运动时间为t秒.
(1)当$t= 2$时,OQ的长度为____；
(2)如图1,设MN,PN分别与直线$y= -\frac {2}{3}x+4$交于点C,D,求证:$MC= NC$；
(3)在(2)的条件下,在运动过程中,设正方形PQMN的对角线交于点E,MP与QD交于点F,如图2,求$OF+EN$的最小值.

答案：

(1)$2$ 【解析】令$y = -\frac{2}{3}x + 4 = 0$,解得$x = 6$,$\therefore OA = 6$.$\because t = 2$,$\therefore AP = PQ = 2$,$\therefore OQ = 6 - 2 - 2 = 2$.
(2)证明:$\because AP = PQ = t$,$\therefore OQ = 6 - 2t$.$\because$四边形$PQMN$是正方形,$\therefore PQ = QM = MN = PN=t$,$\therefore M(6 - 2t,t)$,$N(6 - t,t)$,$C(6-\frac{3}{2}t,t)$,$\therefore MC=(6-\frac{3}{2}t)-(6 - 2t)=\frac{1}{2}t$,$NC=(6 - t)-(6-\frac{3}{2}t)=\frac{1}{2}t$,$\therefore MC = NC$.
(3)解:作长方形$NEFK$,连接$OF$,则$EN = FK$.$\because OF + EN = OF + FK$,$\therefore$当$O$,$F$,$K$三点共线时,$OF + EN = OF + FK$的值最小为$OK$,如图.过点$O$作$OH\perp QN$于点$H$,则$\triangle OQH$是等腰三角形.$\because OQ = 2t - 6$,$\therefore QH = OH=\sqrt{2}t - 3\sqrt{2}$.在等腰直角三角形$PQN$中,$\because PQ = t$,$\therefore QN=\sqrt{2}t$,$\therefore HN = QN - QH=\sqrt{2}t-(\sqrt{2}t - 3\sqrt{2})=3\sqrt{2}$,$\therefore OF + EN$的最小值为$OK = HN = 3\sqrt{2}$.