10. 解：(1)∵点在反比例函数图象上，

∴，

即反比例函数关系式为；

∵点在反比例函数图象上，

∴，

∵点和在一次函数的图象上，

∴，

解得，

∴一次函数关系式为.

(2)当时，一次函数值为2，

∴，

∴.

9. 乙题：

解：(1)因为反比例函数的图象经过点

有，················································································································ 2分

．····················································································································· 3分

所以反比例函数的解析式为，············································································· 4分

(2)当为一、三象限角平分线与反比例函数图像的交点时，

线段最短．············································································································ 5分

将代入，解得，即，．····················· 6分

，··········································································································· 7分

则，··········································································································· 8分

又为反比例函数图像上的任意两点，

由图象特点知，线段无最大值，即．·················································· 9分

8.解：(1)∵反比例函数的图像经过点A(1，3)，

　　　　　　 ∴，即m=-3.

　　　　　　 ∴反比例函数得表达式为.　　　　　　　　　　　　 ……3分

　　　　　　 ∵一次函数y=kx+b的图像经过A(1,-3)、C(0,-4)，

　　　　　　 ∴　 解得

　　　　　　 ∴一次函数的表达式为y=x-4　　　　　　　　　　　　　　 ……3分

(2)由消去y，得x²-4x+3=0.

　 　即(x-1)(x-3)=0.

　　 ∴x=1或x=3.

　　 可得y=-3或y=-1.

于是或

而点A的坐标是(1，-3)，

∴点B的坐标为(3，-1)。　　　　　　　　　　　　　　　　　　　　 ……2分

7. 解：(1)设药物燃烧阶段函数解析式为，由题意得：

························································································································ 2分

．此阶段函数解析式为······································································· 3分

(2)设药物燃烧结束后的函数解析式为，由题意得：

·························································································································· 5分

．此阶段函数解析式为······································································ 6分

(3)当时，得···················································································· 7分

························································································································· 8分

·························································································································· 9分

从消毒开始经过50分钟后学生才可回教室．···························································· 10分

6. 解 (Ⅰ)∵点P(2，2)在反比例函数的图象上，

∴．即． ···································································································· 2分

∴反比例函数的解析式为．

∴当时，． ···························································································· 4分

(Ⅱ)∵当时，；当时，，　 ····················································· 6分

又反比例函数在时值随值的增大而减小， ············································· 7分

∴当时，的取值范围为．································································ 8分

5. (1)证明：分别过点C、D作

垂足为G、H，则

(2)①证明：连结MF，NE

设点M的坐标为，点N的坐标为，

∵点M，N在反比例函数的图象上，

∴，

由(1)中的结论可知：MN∥EF。

②MN∥EF。

4. 解：(1)由题意可知，．

解，得 m＝3．　　　　 ………………………………3分

∴ A(3，4)，B(6，2)；

∴ k＝4×3=12．　　 ……………………………4分

(2)存在两种情况，如图：　

①当M点在x轴的正半轴上，N点在y轴的正半轴

上时，设M₁点坐标为(x₁，0)，N₁点坐标为(0，y₁)．

∵ 四边形AN₁M₁B为平行四边形，

∴ 线段N₁M₁可看作由线段AB向左平移3个单位，

再向下平移2个单位得到的(也可看作向下平移2个单位，再向左平移3个单位得到的)．

由(1)知A点坐标为(3，4)，B点坐标为(6，2)，

∴ N₁点坐标为(0，4－2)，即N₁(0，2)；　　　 ………………………………5分

M₁点坐标为(6－3，0)，即M₁(3，0)．　　　 ………………………………6分

设直线M₁N₁的函数表达式为，把x＝3，y＝0代入，解得．

∴ 直线M₁N₁的函数表达式为． ……………………………………8分

②当M点在x轴的负半轴上，N点在y轴的负半轴上时，设M₂点坐标为(x₂，0)，N₂点坐标为(0，y₂)．　

∵ AB∥N₁M₁，AB∥M₂N₂，AB＝N₁M₁，AB＝M₂N₂，

∴ N₁M₁∥M₂N₂，N₁M₁＝M₂N₂．　　

∴ 线段M₂N₂与线段N₁M₁关于原点O成中心对称．　　　

∴ M₂点坐标为(-3，0)，N₂点坐标为(0，-2)．　　 ………………………9分

设直线M₂N₂的函数表达式为，把x＝-3，y＝0代入，解得，

∴ 直线M₂N₂的函数表达式为． 　　

所以，直线MN的函数表达式为或．　 ………………11分

(3)选做题：(9，2)，(4，5)．　 ………………………………………………2分

3. 解：(1) ………(每个点坐标写对各得2分)………………………4分

(2) ∵　　　 ∴…1分

　 ∴ …………………1分

　　　　 　∴ …………………2分

　(3)　 ① ∵

∴相应B点的坐标是 …………………………………………1分

∴. ………………………………………………………………1分

　② 能　 ……………………………………………………………………1分

　　 当时，相应,点的坐标分别是，

经经验：它们都在的图象上

　…………………………………………………………………1分

2. 解：(1)(-4，-2)；(-m,-)

(2) ①由于双曲线是关于原点成中心对称的，所以OP=OQ,OA=OB,所以四边形APBQ一定是平行四边形

②可能是矩形，mn=k即可

不可能是正方形，因为Op不能与OA垂直。

解：(1)作BE⊥OA，

∴ΔAOB是等边三角形

∴BE=OB·sin60^o=，

∴B(,2)

∵A(0,4),设AB的解析式为,所以,解得,的以直线AB的解析式为

(2)由旋转知，AP=AD, ∠PAD=60^o,

∴ΔAPD是等边三角形，PD=PA=

1. 证明：(1)分别过点C，D，作CG⊥AB，DH⊥AB，

垂足为G，H，则∠CGA＝∠DHB＝90°．……1分

∴ CG∥DH．　　

∵ △ABC与△ABD的面积相等，　

∴ CG＝DH．　 　…………………………2分

∴ 四边形CGHD为平行四边形．　

∴ AB∥CD．　 ……………………………3分

(2)①证明：连结MF，NE．　 …………………4分

设点M的坐标为(x₁，y₁)，点N的坐标为(x₂，y₂)．

∵ 点M，N在反比例函数(k＞0)的图象上，

∴ ，．　

∵ ME⊥y轴，NF⊥x轴，　

∴ OE＝y₁，OF＝x₂．

∴ S_△EFM＝，　　 ………………5分

S_△EFN＝．　　 ………………6分

∴S_△EFM ＝S_△EFN． ……

由(1)中的结论可知：MN∥EF．　 ………8分

② MN∥EF．　 　　　　…………………10分

(若学生使用其他方法，只要解法正确，皆给分．