如图，在平面直角坐标系中，抛物线经过A(?1，0)，B(3，0)，C(0，-1)三点. （1）求该抛物线的表达式；
（2）点Q在y轴上，点P在抛物线上，要使以点Q、P、A、B为顶点的四边形是平行四边形，求所有满足条件的点P的坐标.
25.（本题满分12分） 问题探究
（1）请你在图①中作一条直线，使它将矩形ABCD分成面积相等的两部分； ．．
（2）如图②，点M是矩形ABCD内一定点.请你在图②中过点M作一条直线，使它将矩形ABCD分成面积相等的两部分. 问题解决
（3）如图③，在平面直角坐标系中，直角梯形OBCD是某市将要筹建的高新技术开发区用地示意图，其中DC∥OB，OB?6，BC?4，CD?4.开发区综合服务管理委员会（其占地面积不计）设在点P(4，，并且使这条路所2)处.为了方便驻区单位，准备过点P修一条笔直的道路（路的宽度不计）在的直线l将直角梯形OBCD分成面积相等的两部分.你认为直线l是否存在？若存在，求出直线l的表达式；若不存在，请说明理由.
2010年陕西省初中毕业学业考试
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一、选择题（共10小题，每小题3分，计30分）
题号 A卷答案 1 C 2 B 3 B 4 D 5 A 6 C 7 A 8 A 9 D 10 C 二、填空题（共6小题，每小题3分，计18分） 11. ?2 12.x?0或x?4 13.?ACD??B(?ADC??ACB或14. 0.4 15.?12 16. 18
三、解答题（共9小题，计72分）（以下给出了各题的一种解法及评分参考，其它符合题意的解法请参照相应题的解答赋分） 17.解：原式=
m(m?n)(m?n)(m?n)2ADAC?ACAB之一亦可)
?n(m?n)(m?n)(m?n)2?2mn(m?n)(m?n)
=
m?mn?nm?n?2mn(m?n)(m?n)m?2mn?n22 ··························································· （3分）
=
(m?n)(m?n)(m?n)2
= =
(m?n)(m?n) ············································································ （4分）
m?nm?n ·························································································· （5分）
18.证明：在正方形ABEF和正方形BCMN中，
AB?BE?EF，BC?BN，?FEN??EBC?90°.········································· （2?AB?2BC，
?EN?BC. ········································································································ （4?△FEN≌△EBC分）
分）
. ··························································································· （5分）
分）
?FN?EC. ········································································································ （6
19.解：（1）如图所示. ···················································································· （2分）
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（2）24?161 600············································································ （5?20%?1.8.·分）
?该县常住居民利用“五一”期间出游采集发展信息的人数约为1.8万人. （6分）
（3）略.（只要谈出合理、健康、积极的感想即可给分） ······················· （7分） 20.解：过点P作PH⊥AB，垂足为H．则?APH?30°，?BPH?43°. （1分） 在Rt△APH中，
AH?100，PH?AP·cos30°?1003． ··············· （4
分）
在Rt△PBH中，
·············· （6BH?PH·tan43 ≈1003?0.933≈161.60．?AB?AH?BH≈100?161.60≈262．
分）
答：码头A与亭子B之间的距离约为262米. ············································· （8分） 21.解：（1）由题意，得批发蒜薹3x吨，储藏后销售（200?4x）吨， ··· （2分） 则y?3x·(3 000?700)?x·(4 500?1 000)?(200?4x)·(5 500?1 200)
=?6 800x?860 000. ················································································· （4分） （2）由题意，得200?4x≤80.解之，得x≥30． ··········································· （6分） ?y??6 800x?860 000，?6 800?0．?y的值随x的值增大而减小.
时，y最大值??6 800?30?860 000?656 000．
元. ··················· （8分）
?当x?30?该生产基地按计划全部售完蒜薹的最大利润为656 00022.解：（1）游戏所有可能出现的结果如下表：
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···························································································································· （4分） 从上表可以看出，一次游戏共有20种等可能结果，其中两数和为偶数的共有8种.将参 加联欢会的某位同学即兴表演节目记为事件A，
?P(A)?P（两数和为偶数）=
820?25． ························································· （6
分）
（2）?50?25?20（人）.
?估计本次联欢会上有20名同学即兴表演节目. ········································ （8分）
23.解：（1）?DE垂直平分AC，
??DEC?90°．
?DC为△DEC?DC外接圆的直径.
的中点O即为圆心.
·················································································（1分） 连接OE.
又知BE是⊙O的切线，
??EBO??BOE?90°． ··················································································· （2
分）
在Rt△ABC中，E是斜边AC的中点，
?BE?EC． ??EBC??C.
又??BOE?2?C,
??C?2?C?90°.
??C?30°. ······································································································· （4
分）
（2）在Rt△ABC中，AC??EC?
AB?BC22?5. 12AC?52． ··························································································· （6
9
分）
??ABC??DEC?90°，
?△ABC∽△DEC． ?AC?DC54BC ．EC ?DC?.
58C ?△DE外接圆的半径为. ···································································· （8分）
24.解：（1）设该抛物线的表达式为y?ax2?bx?c．根据题意，得
1?a?，?3?a?b?c?0，?2??b??，9a?3b?c?0，解之，得 ································································· （2分） ??3??c??1．??c??1．???所求抛物线的表达式为y?13x?223x?1．
（2）①当AB为边时，只要PQ∥AB，且PQ?AB?4即可. 又知点Q在y轴上，?点P的横坐标为4或?4. 这时，符合条件的点P有两个，分别记为P1，P2． 而当x?4时，y???5?3?53；当x??4时，y?7．
7)．此时P1?4，?，P2(?4， ·················································································· （7分）
②
当AB为对角线时，只要线段PQ与线段AB互相平分即可.
又知点Q在y轴上，且线段AB中点的横坐标为1，
?点P的横坐标为2.
这时，符合条件的点P只有一个，记为P3.
?1). 而当x?2时，y??1．此时P3(2，7)，P3(2，．1) 综上，满足条件的点P为P1?4，?，P2(?4，······························· （10分）
??5?3?25.解：（1）如图①，作直线DB，直线DB即为所求.（所求直线不唯一，只要过矩形 对称中心的直线均可） ···················································································· （2分）
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（2）如图②，连接AC、DB交于点P，则点P为矩形ABCD的对称中心.作直线MP， 直线MP即为所求. ·························································································· （5分） （3）如图③，存在符合条件的直线l. ·························································· （6分） 过点D作DA⊥OB于点A，
则点P(4，························································· （6分） 2)为矩形ABCD的对称中心. ·
?过点P的直线只要平分△DOA的面积即可.
易知，在OD边上必存在点H，使得直线PH将△DOA面积平分. 从而，直线PH平分梯形OBCD的面积.
即直线PH为所求直线l. ·················································································· （9分） 设直线PH的表达式为y?kx?b，且点P(4， 2)，?2?4k?b．即b?2?4k． ?y?kx?2?4k．?直线OD的表达式为y?2x．
2?4k?x?，??y?kx?2?4k，?2?k解之，得? ???y?2x．?y?4?8k．?2?k??点H的坐标为??2?4k4?8，2?k2?k?? ?．??PH与线段AD的交点F的坐标为(2，2?2k)，
2k?2．??4?k?1． ?0??
?S△DHF?12?4?2?k4???2k·?2?2???2?k???211 4??2．2 解之，得k?8 ?b??13?3??13?3．k?不合题意，舍去? ???22???2．13
?直线l的表达式为y?13?32x?8?213． ················································ （12分）
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