小题1:对于 ,令x=0,得y=4,即B(0,4);… 令y=0,即 ,解得:x1 = —2,x2 = 4,即A(4,0) 设直线AB的解析式为y =" kx" + b, 把A(4,0),B(0,4)分别代入上式,得
,解得:k = —1,b = 4, ∴直线AB的解析式为y = —x + 4。 小题1:当点P(x,y)在直线AB上时,由x = —x + 4,得:x = 2, 当点Q在直线AB上时,依题意可知Q( , ),由 ,得:x = 4, ∴若正方形PEQF与直线AB有公共点,则x的取值范围为2≤x≤4; 小题1:当点E(x, )在直线AB上时, ,解得 , ①当 时,直线AB分别与PE、PF交于点C、D,此时PC = x—(—x+4) = 2x—4, ∵ PD = PC, ∴ S△PCD =![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070418-78077.png) ∴![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070419-21783.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070419-16305.png)
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070419-88303.png) ∵ , ∴当 时,![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070420-81353.png) ②当 时,直线AB分别与QE、QF交于点M、N,此时,
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070420-54993.png) ∵ QM = QN, ∴ S△QMN=![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070420-32804.png) 即 , 其中,当 时,![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070421-69021.png) 综合①、②,当 时,![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019070420-81353.png) |