2013-2014年上海市浦东新区初二(上)期末数学试卷含参考答案

n的代数式表示点B的坐标；

（3）在第（2）小题的条件下，求的值．

【解答】解：（1）过A作AC⊥OB，交x轴于点C， ∵OA=AB，∠OAB=90°， ∴△AOB为等腰直角三角形， ∴AC=OC=BC=OB=3， ∴A（3，3），

将x=3，y=3代入反比例解析式得：3=，即k=9， 则反比例解析式为y=；

（2）过A作AE⊥x轴，过B作BD⊥AE， ∵∠OAB=90°， ∴∠OAE+∠BAD=90°， ∵∠AOE+∠OAE=90°， ∴∠BAD=∠AOE， 在△AOE和△BAD中，

，

∴△AOE≌△BAD（AAS）， ∴AE=BD=n，OE=AD=m，

∴DE=AE﹣AD=n﹣m，OE+BD=m+n， 则B（m+n，n﹣m）；

第17页（共19页）

（3）由A与B都在反比例图象上，得到mn=（m+n）（n﹣m）， 整理得：n2﹣m2=mn，即（）2+﹣1=0， 这里a=1，b=1，c=﹣1， ∵△=1+4=5， ∴=

，

∵A（m，n）在第一象限， ∴m＞0，n＞0， 则=

．

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