《2025年名校题库八年级数学上册北师大版》

7. (西川)定义:对于平面直角坐标系xOy中的不在同一条直线上的三点P,M,N,若满足点M绕点P逆时针旋转$90^{\circ }$后恰好与点N重合,则称点N为点M关于点P的“逆旋点”.请根据以上定义,完成下列填空:
(1)若点M在直线$y= 3x+3$上,点P与原点O重合,且点M关于点P的“逆旋点”N恰好在坐标轴上,则点N的坐标为____；
(2)如图,已知点A的坐标为$(\frac {3}{2},0)$,C是y轴上的动点,点B是点A关于点C的“逆旋点”,连接OB,AB,则$OB+AB$的最小值是____.

答案：

(1)$(-3,0)$或$(0,-1)$
(2)$\frac{3}{2}\sqrt{5}$ 【解析】
(1)①当点$N$落在$x$轴上时,$OM\perp x$轴,$\therefore$点$M$在$y$轴上,当$x = 0$时,$y = 3x + 3 = 3$,$\therefore M(0,3)$,$\therefore N(-3,0)$.②当点$N$落在$y$轴上时,$OM\perp y$轴,$\therefore$点$M$在$x$轴上,当$y = 0$时,$3x + 3 = 0$,解得$x = -1$,$\therefore M(-1,0)$,$\therefore N(0,-1)$.综上所述,点$N$的坐标为$(-3,0)$或$(0,-1)$.
(2)如图1,过点$B$作$BH\perp y$轴于点$H$.设点$C$的坐标为$(0,m)$.$\because\angle ACB = 90^{\circ}$,$\therefore\angle ACO+\angle BCH = 90^{\circ}$.$\because\angle ACO+\angle OAC = 90^{\circ}$,$\therefore\angle BCH=\angle OAC$.又$\because\angle BHC=\angle COA = 90^{\circ}$,$BC = CA$,$\therefore\triangle BCH\cong\triangle CAO(AAS)$,$\therefore OC = HB=m$,$OA = HC=\frac{3}{2}$,$\therefore B(m,\frac{3}{2}+m)$,$\therefore BO+BA=\sqrt{m^{2}+(m+\frac{3}{2})^{2}}+\sqrt{(m-\frac{3}{2})^{2}+(m+\frac{3}{2})^{2}}$,相当于在直线$y = x$上寻找一点$P(m,m)$,使得点$P$到点$M(0,-\frac{3}{2})$,$N(\frac{3}{2},-\frac{3}{2})$的距离和最小.如图2,作点$M$关于直线$y = x$的对称点$M'(-\frac{3}{2},0)$,连接$PM'$,$M'N$,$\therefore M'N=\sqrt{(\frac{3}{2}+\frac{3}{2})^{2}+(\frac{3}{2})^{2}}=\frac{3}{2}\sqrt{5}$.$\because PM + PN=PM'+PN\geqslant M'N$,$\therefore PM + PN$的最小值为$\frac{3}{2}\sqrt{5}$,$\therefore OB + AB$的最小值为$\frac{3}{2}\sqrt{5}$.

8. (树德实验)如图,在平面直角坐标系中,已知点$A(8,4)$,$AB⊥y$轴于点B,$AC⊥x$轴于点C,直线$y= x$交AB于点D.
(1)直接写出B,C,D三点的坐标;
(2)如图1,若E为OD延长线上一动点,记点E的横坐标为a,$△BCE$的面积为S,求S与a之间的关系式;
(3)如图2,在(2)的条件下,当$S= 20$时,过点E作$EF⊥AB$于点F,G,H分别为AC,CB上的动点,求$FG+GH$的最小值.

答案：
解:
(1)$B(0,4)$,$C(8,0)$,$D(4,4)$.
(2)由题意,得$E(a,a)$,$\therefore S = S_{\triangle BDE}+S_{\triangle CDE}-S_{\triangle BDC}=4a×\frac{1}{2}+8a×\frac{1}{2}-8×4×\frac{1}{2}=6a - 16$.
(3)当$S = 20$时,$20 = 6a - 16$,解得$a = 6$,$\therefore E(6,6)$.$\because EF\perp AB$于点$F$,$\therefore F(6,4)$.如图,作点$F$关于直线$AC$的对称点$F'(10,4)$,作$F'H\perp BC$于点$H$,交$AC$于点$G$,此时$FG + GH$的值最小.易求得直线$BC$的表达式为$y = -\frac{1}{2}x + 4$,直线$F'H$的表达式为$y = 2x - 16$.由$y = -\frac{1}{2}x + 4$和$y = 2x - 16$,可求得$H(8,0)$,$\therefore F'H = 2\sqrt{5}$,$\therefore FG+GH$的最小值为$2\sqrt{5}$.