如图，在平面直角坐标系中有Rt△ABC，∠A=90°，AB=AC，A（﹣2，0）、B（0，d）、C（﹣3，2）...

问题详情：

如图，在平面直角坐标系中有Rt△ABC，∠A=90°，AB=AC，A（﹣2，0）、B（0，d）、C（﹣3，2）．                                                        

（1）求d的值；                                                                                

（2）将△ABC沿x轴的正方向平移a个单位，在第一象限内B、C两点的对应点B′C′正好落在某反比例函数图象上．请求出这个反比例函数和此时直线B′C′的解析式；                                      

（3）在（2）的条件下，直线B′C′交y轴于点G，作C′M⊥x轴于M．P是线段B′C′上的一点，若△PMC′和△PBB′面积相等，求点P坐标．                                                                              

【回答】

解：（1）作CN⊥x轴于点N．                                                           

在Rt△CNA和Rt△AOB中，                                                               

，                                                                                         

∴Rt△CNA≌Rt△AOB（HL），                                                          

则BO=AN=3﹣2=1，                                                                          

∴d=1；                                                                                             

（2）设反比例函数为y=，点C′和B′在该比例函数图象上，                          

设C′（a，2），则B′（a+3，1）                                                          

把点C′和B′的坐标分别代入y=，得k=2a；k=a+3，                                       

∴2a=a+3，a=3，                                                                               

则k=6，反比例函数解析式为y=．                                                    

得点C′（3，2）；B′（6，1）；                                                           

设直线C′B′的解析式为y=ax+b，把C′、B′两点坐标代入得，             

解得：；                                                                              

∴直线C′B′的解析式为：y=﹣；                                                 

（3）连结BB′                                                                                    

∵B（0，1），B′（6，1），                                                                

∴BB′∥x轴，                                                                                    

设P（m，），作PQ⊥C′M，PH⊥BB′                                                

∴S△PC’M=×PQ×C′M=×（m﹣3）×2=m﹣3                                       

S△PBB’=×PH×BB′=×（）×6=﹣m+6                                           

∴m﹣3=﹣m+6                                                                                  

∴m=                                                                                               

∴P（，）．                                                                                 

知识点：反比例函数

题型：综合题