中考数学压轴题100题精选（41-50题）答案
【041】（1）如图 （3分） y（千米）
150
100
50
0 1 2 -1
D B E A C 6 7 8 x（小时） 5 3 4 （2）2次······························································································································· （5分） （3）如图，设直线AB的解析式为y?k1x?b1，
0)B(6，150)， 图象过A(4，，?4k1?b1?0，?k1?75，y?75x?300．① ···················································· （7分） ?????6k1?b1?150.?b1??300.设直线CD的解析式为y?k2x?b2， 图象过C(7，，0)D(5150)，，
?7k2?b2?0，?k2??75，············································· （7分） ?y??75x?525．② ·????5k?b?150.b?525.?22?2解由①、②组成的方程组得??x?5.5，
?y?112.5.?最后一次相遇时距离乌鲁木齐市的距离为112.5千米． （12分） 【042】解：（1）∵点D是OA的中点，∴OD?2，∴OD?OC． 又∵OP是?COD的角平分线，∴?POC??POD?45°， ∴△POC≌△POD，∴PC?PD． ··················································································· 3分 （2）过点B作?AOC的平分线的垂线，垂足为P，点P即为所求．
易知点F的坐标为（2，2），故BF?2，作PM⊥BF， 1∵△PBF是等腰直角三角形，∴PM?BF?1，
2∴点P的坐标为（3，3）．
2y CP (0，2) F M B A(4，0) x ∵抛物线经过原点，∴设抛物线的解析式为y?ax?bx． O E D ?9a?3b?3?a?13)和点D(2，0)，∴有?又∵抛物线经过点P(3， 解得?
?4a?2b?0?b??2∴抛物线的解析式为y?x2?2x． ·························································································· 7分 （3）由等腰直角三角形的对称性知D点关于?AOC的平分线的对称点即为C点．
连接EC，它与?AOC的平分线的交点即为所求的P点（因为PE?PD?EC，而两点之间线段最短），此时△PED的周长最小． ∵抛物线y?x2?2x的顶点E的坐标(1，?1)，C点的坐标(0，2)，
?k?b??1?k??3CE设所在直线的解析式为y?kx?b，则有?，解得?．
b?2b?2??∴CE所在直线的解析式为y??3x?2．
1?x???y??3x?2??11?2点P满足?，解得?，故点P的坐标为?，?．
?22??y?x?y?1??2△PED的周长即是CE?DE?10?2．
2)． ·（4）存在点P，使?CPN?90°．其坐标是?，?或(2，····································· 14分
【043】解（Ⅰ）
?11??22?y1?x，y2?x2?bx?c，y1?y2?0，
······································································································· 1分 ?x2??b?1?x?c?0.·将??，??21312分别代入x??b?1?x?c?0，得 2211?1??1??b?1??c?0，?b?1??c?0， ????????32?3??2?解得b?1151，c?.?函数y2的解析式为y2?x2?x?． ······································ 3分 6666（Ⅱ）由已知，得AB?2，设△ABM的高为h， 6?S△ABM?1121AB·h?h?3，即2h?.
1442121211511. 2h，由T?t2?t?，得?t2?t??6666144根据题意，t?T?2当t?5115t???时，解得t1?t2?； 6614412当t?25115?25?2t??时，解得t3?. ，t4?661441212?t的值为
（
Ⅲ
55?25?2. ·························································································· 6分 ，，121212）
由
已
知
，
得
???2?b??c，???2?b??c，T?t2?bt?c.
?T????t????t???b?T????t????t???b?，
，
??????2?b??c????2?b??c?，化简得??????????b?1??0.
0?????1，得????0， ?????b?1?0.
有??b?1???0，??b?1???0. 又0?t?1，?t???b?0，t???b?0，
?当0?t≤a时，T≤?≤?；当??t≤?时，??T≤?；
当??t?1时，????T. ····································································································· 10分 【044】(1) 配方,得y=
取x=0代入y=
12
(x–2) –1，∴抛物线的对称轴为直线x=2，顶点为P(2，–1) ． 212
x –2x＋1，得y=1，∴点A的坐标是(0，1)．由抛物线的对称性知，点A(0，21)与点B关于直线x=2对称，∴点B的坐标是(4，1)． 2分
设直线l的解析式为y=kx＋b(k≠0)，将B、P的坐标代入，有 ?1?4k?b,?k?1,解得∴直线l的解析式为y=x–3．3分 ???1?2k?b,b??3.??(2) 连结AD交O′C于点E，∵ 点D由点A沿O′C翻折后得到，∴ O′C垂直平分AD． 由(1)知，点C的坐标为(0，–3)，∴ 在Rt△AO′C中，O′A=2，AC=4，∴ O′C=25．
1148据面积关系，有 ×O′C×AE=×O′A×CA，∴ AE=5，AD=2AE=5．
2255AFDFAD作DF⊥AB于F，易证Rt△ADF∽Rt△CO′A，∴， ??ACO?AO?CAD16AD8∴ AF=·AC=，DF=·O′A=，5分
O?C5O?C583又 ∵OA=1，∴点D的纵坐标为1–= –，
55163∴ 点D的坐标为(，–)．
55