如图,在平面直角坐标系中,▱ABCD的顶点A、B、D的坐标分别为(2,0

如图,在平面直角坐标系中,ABCD的顶点ABD的坐标分别为(20)、(60)、(03),顶点C在函数y=x0)的图象上.                                                                               

1)求k的值.                                                                                

2)将ABCD向上平移,当点B恰好落在函数y=x0)的图象上时,                

求平移的距离;                                                                              

CD与函数y=x0)图象的交点坐标.                                             

                                                                       

                                                                                                       

                                                                          

答案

【分析】1)根据平行四边形的性质求出点C坐标,代入函数解析式中求出k                

2根据平移的性质,得到点B的横坐标不变是6,从而确定出平移距离即可;                   

先确定出点D平移后的坐标,由平移的性质确定出交点坐标.                    

【解答】解:(1)在平行四边形ABCD中,A20),B60),D03),                   

CD=AB=4CDAB                                                                         

C43),                                                                              

C在函数y=x0)的图象上.                                                   

k=4×3=12                                                                                  

2由(1)有,k=12                                                                     

函数的解析式为y=x0),                                                        

ABCD向上平移,                                                                        

B的横坐标不变仍是6                                                                   

平移后点B在函数y=的图象上,                                                    

此时点B的纵坐标为=2                                                                

平移的距离为2个单位,                                                                     

知,平移后点B坐标为(62),                                                

平移后点D的坐标为(05),                                                          

此时CD与函数y=的图象的交点的纵坐标是5,而当y=5时,x=                 

CD与函数y=的图象的交点的坐标是(5).                                 

【点评】此题是反比例函数综合题,主要考查了待定系数法,平行四边形的性质,平移的性质,解本题的关键是掌握平移的性质的同时灵活运用.

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