广东省梅州市2009年初中毕业生学业考试数学试题（含答案）

21．本题满分 8 分． （1）解：令x?0，得y?33233·················································· 1分 3，得点C(0，3)． ·

令y?0，得?x?2x?3?0，解得x1??1，x2?3，

∴A(?1，························································································· 3分 0)，B(3，0)． ·（2）法一：证明：因为AC2?12?(3)2?4，

BC2y M2 C N M1 B x

······················ 4分 ?3?(3)?12，AB?16， ·

222222∴AB?AC?BC，············································ 5分 ∴△ABC是直角三角形．········································ 6分 法二：因为OC?3，OA?1，OB?3，

A O M3 21题图

∴OC2?OA?OB， ····································································································· 4分 ∴

OCOA?OBOC，又?AOC??COB，

∴Rt△AOC∽Rt△COB． ······················································································· 5分 ∴?ACO??OBC，?OCB??OBC?90°， ∴?ACO??OCB?90°，

∴?ACB?90°， 即△ABC是直角三角形． ·················································6 分 （3）M1(4，3)，M2(?4，3)，M3(2，?22．本题满分 10 分． （1）①

65（只写出一个给1分，写出2个，得1.53)．

分） ······························································· 8分

D O A E G F 22题图

B ······························································· 2分

C ②法一：在矩形ABCD中，AD?BC，

?ADE??BCE，又CE?DE，

∴△ADE≌△BCE， ············································ 3分

得AE?BE，?EAB??EBA，

连OF，则OF?OA， ∴?OAF??OFA，

?OFA??EBA， ∴OF∥EB， ············································································4 分

∵FG⊥BE， ∴FG⊥OF， ∴FG是⊙O的切线 ························································································· 6分 （法二：提示：连EF，DF，证四边形DFBE是平行四边形．参照法一给分．） （2）法一：若BE能与⊙O相切， ∵AE是⊙O的直径， ∴AE⊥BE，则?DEA??BEC?90°，

又?EBC??BEC?90°， ∴?DEA??EBC，

∴Rt△ADE∽Rt△ECB，

∴

ADEC?DEBC，设DE?x，则EC?5?x，AD?BC?3，得

35?x?x3，

整理得x2?5x?9?0． ·······························································································8 分 ∵b2?4ac?25?36??11?0， ∴该方程无实数根．

∴点E不存在，BE不能与⊙O相切． ·······································10分 法二： 若BE能与⊙O相切，因AE是⊙O的直径，则AE⊥BE，?AEB?90°， 设DE?x，则EC?5?x，由勾股定理得：AE2?EB2?AB2，

即(9?x2)?[(5?x)2?9]?25， 整理得x2?5x?9?0， ··································· 8分 ∵b2?4ac?25?36??11?0， ∴该方程无实数根．

∴点E不存在，BE不能与⊙O相切． ·······································10分 （法三：本题可以通过判断以AB为直径的圆与DC是否有交点来求解，参照前一解法给分） 23．本题满分 11 分．

（1）y?1?x·············································································································· 2分 （2）∵OP?t，∴Q点的横坐标为①当0?1212t，

12t，

t?1，即0?t?2时，QM?1?∴S△OPQ?1?1?···························································································· 3分 t?1?t?． ·

2?2?1212②当t≥2时，QM?1?t?t?1，

∴S△OPQ?1?1?t?t?1?． 2?2??1?1?t1?t?，0?t?2，??2??2?∴S?? ······················································································ 4分

?1t?1t?1?，t≥2.?????2?2当0?12t?1，即0?t?2时，S?141?1?112t?1?t???(t?1)?， 2?2?44∴当t?1时，S有最大值．······················································································ 6分

（3）由OA?OB?1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ

是以Q为直角顶点的等腰直角三角形，则PQ?QC，所以OQ?QC，又L1∥x轴，则C，

O两点关于直线L对称，所以AC?OA?1，得C(1，···········································7 分 1)． ·

下证?PQC?90°．连CB，则四边形OACB是正方形．

y 法一：（i）当点P在线段OB上，Q在线段AB上 （Q与B、C不重合）时，如图–1．

由对称性，得?BCQ??QOP，?QPO??QOP， ∴ ?QPB??QCB??QPB??QPO?180°，

∴ ?PQC?360°?(?QPB??QCB??PBC)?90°． ············································ 8分 （ii）当点P在线段OB的延长线上，Q在线段AB上时，如图–2，如图–3

∵?QPB??QCB，?1??2， ∴?PQC??PBC?90°． ························ 9分 （iii）当点Q与点B重合时，显然?PQC?90°． 综合（i）（ii）（iii），?PQC?90°．

∴在L1上存在点C(1，··········· 11 分 1)，使得△CPQ是以Q为直角顶点的等腰直角三角形．·

L A Q O 2 1 L A Q C L1

O P B 23题图-1 x

y C L1 y L A 1 C L1 P B 23题图-2 x O B 2 Q P x 23题图-3

法二：由OA?OB?1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ是以Q为直角顶点的等腰直角三角形，则PQ?QC，所以OQ?QC，又L1∥x轴， 则C，

O两点关于直线L对称，所以AC?OA?1，得C(1，··········································7 分 1)． ·

延长MQ与L1交于点N．

（i）如图–4，当点Q在线段AB上（Q与A、B不重合）时，

∵四边形OACB是正方形，

∴四边形OMNA和四边形MNCB都是矩形，△AQN和△QBM都是等腰直角三角形． ∴NC?MB?MQ，NQ?AN?OM，?QNC??QMB?90°． 又∵OM?MP， ∴MP?QN， ∴△QNC≌△QMP， ∴?MPQ??NQC， 又∵?MQP??MPQ?90°， ∴?MQP??NQC?90°．

∴?CQP?90°． ·································································································· 8分 （ii）当点Q与点B重合时，显然?PQC?90°． ········································· 9分 （iii）Q在线段AB的延长线上时，如图–5， ∵?BCQ??MPQ，∠1=∠2 ∴?CQP??CBM?90°

综合（i）（ii）（iii），?PQC?90°．

∴在L1上存在点C(1，······ 11分 1)，使得△CPQ是以Q为直角顶点的等腰直角三角形． ·

y L A Q O O M P B 23题图-4

x B N C L1

y L A 1 y L A Q C L1

O P B 23题图-1 x C L1 2 Q P x 23题图-5

法三：由OA?OB?1，所以△OAB是等腰直角三角形，若在L1上存在点C，使得△CPQ是以Q为直角顶点的等腰直角三角形，则PQ?QC，所以OQ?QC，又L1∥x轴， 则C，O两点关于直线L对称，所以AC?OA?1，得C(1，····················· 9分 1)． ·

连PC，∵PB?|1?t|，OM?12t，MQ?1?t2，

∴PC2?PB2?BC2?(1?t)2?1?t2?2t?2，

t?t?t??2?MQ?????1????t?1．

2?2?2??222OQ?OP?CQ?OM2222∴PC2?OP2?QC2，∴?CQP?90°．··································································10分 ∴在L1上存在点C(1，·········· 11分 1)，使得△CPQ是以Q为直角顶点的等腰直角三角形． ·