?A(?1，0)，B(3，0)，C(0，?3)．
又y?(x?1)?4，?顶点M(1····························································· 5分 ，?4)． ·容易求得直线CM的表达式是y??x?3． 在y??x?3中，令y?0，得x??3．
·············································································· 6分 ?N(?3，0)，?AN?2． ·在y?x?2x?3中，令y??3，得x1?0，x2?2．
22?CP?2，?AN?CP．
···························· 8分 ?3)． ·AN∥CP，?四边形ANCP为平行四边形，此时P(2，（3）△AEF是等腰直角三角形．
理由：在y??x?3中，令x?0，得y?3，令y?0，得x?3．
?直线y??x?3与坐标轴的交点是D(0，3)，B(3，0)．
····································································· 9分 ?OD?OB，??OBD?45°． ·又
点C(0，·········································· 10分 ?3)，?OB?OC．??OBC?45°． ·
由图知?AEF??ABF?45°，?AFE??ABE?45°． ··································· 11分
··························· 12分 ??EAF?90°，且AE?AF．?△AEF是等腰直角三角形． ·（4）当点E是直线y??x?3上任意一点时，（3）中的结论成立． 14分
2009年烟台市中考数学试题（含答案）
