《2026年金考卷中考45套汇编数学河北专版》

25. (本小题满分12分)
在平面直角坐标系中,设计了点的两种移动方式:从点$(x,y)$移动到点$(x + 2,y + 1)$称为一次甲方式;从点$(x,y)$移动到点$(x + 1,y + 2)$称为一次乙方式.
例 点$P$从原点$O$出发连续移动2次:若都按甲方式,最终移动到点$M(4,2)$;若都按乙方式,最终移动到点$N(2,4)$;若按1次甲方式和1次乙方式,最终移动到点$E(3,3)$.
(1)设直线$l_{1}$经过上例中的点$M,N$,求$l_{1}$的解析式;并直接写出将$l_{1}$向上平移9个单位长度得到的直线$l_{2}$的解析式;
(2)点$P$从原点$O$出发连续移动10次,每次移动按甲方式或乙方式,最终移动到点$Q(x,y)$.其中,按甲方式移动了$m$次.
①用含$m$的式子分别表示$x,y$;
②请说明:无论$m$怎样变化,点$Q$都在一条确定的直线上.设这条直线为$l_{3}$,在图中直

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画出$l_{3}$的图象;
(3)在(1)和(2)中的直线$l_{1},l_{2},l_{3}$上分别有一个动点$A,B,C$,横坐标依次为$a,b,c$,若$A,B,C$三点始终在一条直线上,直

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写出此时$a,b,c$之间的关系式.

答案：
25
(1)设直线$l_{1}$的解析式为$y = kx + b$，把$(4,2)$，$(2,4)$分别代入，得$\begin{cases}4k + b = 2,\\2k + b = 4,\end{cases}$解得$\begin{cases}k = -1,\\b = 6,\end{cases}$$\therefore$直线$l_{1}$的解析式为$y = -x + 6$. (3分)将直线$l_{1}$向上平移$9$个单位长度得到的直线$l_{2}$的解析式为$y = -x + 15$. (4分)
(2)①$\because$点$P$从原点$O$出发连续移动$10$次，按照甲方式移动了$m$次，$\therefore$点$P$按照乙方式移动了$(10 - m)$次，$\therefore x = 2m + 1 × (10 - m) = m + 10$，$y = m + 2 × (10 - m) = -m + 20$. (6分)
②由①知，$y = -m + 20$，$x = m + 10$，$\therefore m = x - 10$，$\therefore y = -(x - 10) + 20 = -x + 30$，$\therefore$无论$m$怎么变化，点$Q$都在直线$y = -x + 30$上. (8分)
直线$l_{3}$的图象如图所示. (10分)

(3)a,b,c之间的关系式为5a + 3c = 8b. (12分)
解法提示：$\because$点$A,B,C$的横坐标分别为$a,b,c$，且分别在直线$l_{1},l_{2},l_{3}$上，$\therefore A(a, -a + 6)$，$B(b, -b + 15)$，$C(c, -c + 30)$.设直线$AB$的解析式为$y = mx + n$，把点$A,B$的坐标分别代入，得$\begin{cases}ma + n = -a + 6,\\mb + n = -b + 15,\end{cases}$解得$\begin{cases}m = -1 + \frac{9}{b - a},\\n = 6 - \frac{9a}{b - a}\end{cases}$，$\therefore$直线$AB$的解析式为$y = (-1 + \frac{9}{b - a})x + 6 - \frac{9a}{b - a}$.
$\because A,B,C$三点始终在一条直线上，$\therefore c(-1 + \frac{9}{b - a}) + 6 - \frac{9a}{b - a} = -c + 30$，整理，得$5a + 3c = 8b$.

26. (本小题满分13分)
如图(1)和图(2),平面上,四边形$ABCD$中,$AB = 8,BC = 2\sqrt{11},CD = 12,DA = 6,\angle A = 90^{\circ}$,点$M$在$AD$边上,且$DM = 2$.将线段$MA$绕点$M$顺时针旋转$n^{\circ}(0 ＜ n\leqslant180)$到$MA'$,$\angle A'MA$的平分线$MP$所在直线交折线$AB—BC$于点$P$,设点$P$在该折线上运动的路径长为$x(x ＞ 0)$,连接$A'P$.
(1)若点$P$在$AB$上,求证:$A'P = AP$.
(2)如图(2),连接$BD$.
①求$\angle CBD$的度数,并直

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写出当$n = 180$时,$x$的值;
②若点$P$到$BD$的距离为2,求$\tan\angle A'MP$的值.
(3)当$0 ＜ x\leqslant8$时,请直接写出点$A'$到直线$AB$的距离(用含$x$的式子表示).

答案：
26
(1)证明：$\because$将线段$MA$绕点$M$顺时针旋转$n^{\circ}$到$MA^{\prime}$，$\therefore A^{\prime}M = AM$.又$\because \angle A^{\prime}MP = \angle AMP$，$PM = PM$，$\therefore \triangle A^{\prime}MP ≌ \triangle AMP(SAS)$，$\therefore A^{\prime}P = AP$. (3分)
(2)①$\because AB = 8$，$DA = 6$，$\angle A = 90^{\circ}$，$\therefore BD = \sqrt{AB^{2} + AD^{2}} = 10$，$\therefore BC^{2} + BD^{2} = (2\sqrt{11})^{2} + 10^{2} = 144$.又$CD^{2} = 12^{2} = 144$，$\therefore BC^{2} + BD^{2} = CD^{2}$，$\therefore \triangle BCD$是直角三角形，且$\angle CBD = 90^{\circ}$(依据：勾股定理的逆定理).当$n = 180$时，$x = 13$. (7分)
解法提示：当$n = 180$时，如图
(1)，设$PM$交$BD$于点$N$.

$\because PM$平分$\angle A^{\prime}MA$，$\therefore \angle PMA = 90^{\circ}$，$\therefore PM // AB$，$\therefore \triangle DNM \backsim \triangle DBA$，$\therefore \frac{DN}{DB} = \frac{DM}{DA} = \frac{MN}{BA}$，即$\frac{DN}{10} = \frac{2}{6} = \frac{MN}{8}$，$\therefore DN = \frac{10}{3}$，$MN = \frac{8}{3}$，$\therefore BN = BD - DN = \frac{20}{3}$.
$\because \angle PBN = \angle NMD = 90^{\circ}$，$\angle PNB = \angle DNM$，$\therefore \triangle PBN \backsim \triangle DMN$，$\therefore \frac{PB}{DM} = \frac{BN}{MN}$，即$\frac{PB}{2} = \frac{\frac{20}{3}}{\frac{8}{3}}$，解得$PB = 5$，$\therefore x = AB + PB = 8 + 5 = 13$.
②分点$P$在$AB$上和点$P$在$BC$上两种情况讨论.
a.当点$P$在$AB$上时，如图
(2)，过点$P$作$PQ \perp BD$于点$Q$，则$PQ = 2$.

$\because \sin \angle DBA = \frac{AD}{BD} = \frac{6}{10} = \frac{3}{5}$，$\therefore BP = \frac{PQ}{\sin \angle DBA} = \frac{2}{\frac{3}{5}} = \frac{10}{3}$，$\therefore AP = AB - BP = 8 - \frac{10}{3} = \frac{14}{3}$，$\therefore \tan \angle A^{\prime}MP = \tan \angle AMP = \frac{AP}{AM} = \frac{\frac{14}{3}}{4} = \frac{7}{6}$. (9分)
b.当点$P$在$BC$上时，如图
(3)，则$PB = 2$.过点$P$作$PK \perp AB$，交$AB$的延长线于点$K$，延长$MP$交$AB$的延长线于点$H$.

$\because \angle PKB = \angle CBD = \angle DAB = 90^{\circ}$，$\therefore \angle KPB = 90^{\circ} - \angle PBK = \angle DBA$，$\therefore \triangle PKB \backsim \triangle BAD$，$\therefore \frac{PK}{BA} = \frac{KB}{AD} = \frac{PB}{BD}$，即$\frac{PK}{8} = \frac{KB}{6} = \frac{2}{10}$，$\therefore PK = \frac{8}{5}$，$BK = \frac{6}{5}$，$\therefore AK = AB + BK = \frac{46}{5}$，$\because PK \perp AB$，$DA \perp AB$，$\therefore PK // DA$，$\therefore \triangle HPK \backsim \triangle HMA$，$\angle KPH = \angle AMP$，$\therefore \frac{HK}{HA} = \frac{PK}{AM}$，即$\frac{HK}{HK + \frac{46}{5}} = \frac{\frac{8}{5}}{4}$，$\therefore HK = \frac{92}{15}$，$\therefore \tan \angle A^{\prime}MP = \tan \angle AMP = \tan \angle KPH = \frac{HK}{PK} = \frac{\frac{92}{15}}{\frac{8}{5}} = \frac{23}{6}$.
综上所述，$\tan \angle A^{\prime}MP$的值为$\frac{7}{6}$或$\frac{23}{6}$. (11分)
(3)$\frac{8x^{2}}{x^{2} + 16}$. (13分)
解法提示：当$0 ＜ x \leq 3$时，点$P$在$AB$上.当$0 ＜ x ＜ 4$时，如图
(4)，过点$A^{\prime}$作$A^{\prime}E \perp AB$于点$E$，过点$M$作$MF \perp EA^{\prime}$，交$EA^{\prime}$的延长线于点$F$，则四边形$AMFE$是矩形，$\therefore AE = FM$，$EF = AM = 4$.

由
(1)知$\triangle A^{\prime}MP ≌ \triangle AMP$，$\therefore \angle PA^{\prime}M = \angle A = 90^{\circ}$，$\therefore \angle PA^{\prime}E + \angle FA^{\prime}M = 90^{\circ}$.又$\angle A^{\prime}MF + \angle FA^{\prime}M = 90^{\circ}$，$\therefore \angle PA^{\prime}E = \angle A^{\prime}MF$.又$\because \angle A^{\prime}EP = \angle MFA^{\prime} = 90^{\circ}$，$\therefore \triangle A^{\prime}PE \backsim \triangle MA^{\prime}F$，$\therefore \frac{A^{\prime}P}{MA^{\prime}} = \frac{PE}{A^{\prime}F} = \frac{A^{\prime}E}{FM}$，设$FM = AE = y$，$A^{\prime}E = h$，则$\frac{x}{4} = \frac{y - x}{4 - h} = \frac{h}{y}$，$\therefore y = \frac{4h}{x}$，$4(y - x) = x(4 - h)$，$\therefore 4(\frac{4h}{x} - x) = x(4 - h)$，整理，得$h = \frac{8x^{2}}{x^{2} + 16}$，即点$A^{\prime}$到直线$AB$的距离为$\frac{8x^{2}}{x^{2} + 16}$.
当$x = 4$时，易知点$A^{\prime}$到$AB$的距离为$4$，也满足$\frac{8x^{2}}{x^{2} + 16}$.
当$4 ＜ x \leq 8$时，如图
(5)，过点$A^{\prime}$作$A^{\prime}Q \perp AB$于点$Q$，过点$M$作$MT \perp A^{\prime}Q$于点$T$，则四边形$TMAQ$是矩形，$\therefore AQ = MT$，$QT = AM = 4$.同理可证$\triangle A^{\prime}PQ \backsim \triangle MA^{\prime}T$，$\therefore \frac{A^{\prime}P}{MA^{\prime}} = \frac{PQ}{A^{\prime}T} = \frac{A^{\prime}Q}{MT}$，设$A^{\prime}Q = m$，$MT = AQ = n$，则$\frac{x}{4} = \frac{x - n}{m - 4} = \frac{m}{n}$，$\therefore n = \frac{4m}{x}$，$4(x - n) = x(m - 4)$，$\therefore 4(x - \frac{4m}{x}) = x(m - 4)$，整理，得$m = \frac{8x^{2}}{x^{2} + 16}$，即点$A^{\prime}$到$AB$的距离为$\frac{8x^{2}}{x^{2} + 16}$.
综上可知，当$0 ＜ x \leq 8$时，点$A^{\prime}$到直线$AB$的距离为$\frac{8x^{2}}{x^{2} + 16}$.