《2025年基础训练大象出版社八年级数学上册人教版》

22. (11分)如图,在平面直角坐标系中,$O$为坐标原点,$A(a,0)$,$B(0,b)$,且$a$,$b满足(a - 4)^{2}+\vert b - 4\vert = 0$.
(1)求点$A$、点$B$的坐标;
(2)动点$P从点O$出发,以每秒$1个单位长度的速度沿y$轴正半轴运动,运动时间为$t\mathrm{s}$,连接$AP$,过点$P作PM\perp AP$,且$PM = PA$,点$M$在第一象限,请用含有$t的式子表示点M$的坐标;
(3)在(2)的条件下,连接$MB$并延长,交$x轴于点Q$,连接$AM$,过点$B作PM的平行线BR$,交$x轴于点R$,当$S_{\triangle MQA}= 28$时,求点$R$的坐标.

答案：

(1) $\because$ a, b满足$(a - 4)^{2} + |b - 4| = 0$,$\therefore a - 4 = 0$, $b - 4 = 0$.$\therefore a = 4$, $b = 4$.$\therefore$ 点A的坐标为$(4, 0)$，点B的坐标为$(0, 4)$.
(2) 如图，过点M作$MC\perp y$轴于点C，则$\angle PCM = \angle AOP = 90^{\circ}$.$\because PM\perp AP$,$\therefore \angle APM = 90^{\circ}$.$\therefore \angle CPM + \angle APO = \angle OAP + \angle APO = 90^{\circ}$.$\therefore \angle CPM = \angle OAP$.在$\triangle CPM$和$\triangle OAP$中，$\left\{\begin{array}{l} \angle PCM = \angle AOP\\ \angle CPM = \angle OAP\\ PM = AP\end{array}\right.$$\therefore \triangle CPM \cong \triangle OAP(AAS)$.$\therefore CM = OP = t$, $CP = OA = 4$.$\therefore CO = CP + OP = 4 + t$.$\therefore$ 点M的坐标为$(t, 4 + t)$.
(3) 由
(1)
(2)，得$CP = OA = OB = 4$.$\therefore CB = OP = t$.$\because CM = t$,$\therefore CB = CM$;$\therefore \triangle BCM$是等腰直角三角形.$\therefore \angle CBM = \angle OBQ = 45^{\circ}$.$\therefore \triangle BOQ$是等腰直角三角形.$\therefore OQ = OB = 4$.$\therefore$ 点Q的坐标为$(-4, 0)$.又$\because$ 点A的坐标为$(4, 0)$，$\therefore AQ = 8$.又$\because$ 点M的坐标为$(t, 4 + t)$，$S_{\triangle MQA} = 28$,$\therefore \frac{1}{2}×8×(4 + t) = 28$.解得$t = 3$.$\therefore OP = 3$.$\because BR// PM$,$\therefore \angle OBR = \angle CPM$.又$\because \angle CPM = \angle OAP$,$\therefore \angle OBR = \angle OAP$.在$\triangle OBR$和$\triangle OAP$中，$\left\{\begin{array}{l} \angle BOR = \angle AOP\\ OB = OA\\ \angle OBR = \angle OAP\end{array}\right.$$\therefore \triangle OBR \cong \triangle OAP(ASA)$.$\therefore OR = OP = 3$.$\therefore$ 点R的坐标为$(-3, 0)$.