053]解:(1)设.把代入.得.··························· 2分 ∴抛物线的解析式为:．顶点的坐标为．······························· 5——青夏教育精英家教网—

053]解:(1)设.把代入.得.··························· 2分 ∴抛物线的解析式为:．顶点的坐标为．······························· 5分 (2)设直线解析式为:().把两点坐标代入. 得解得．∴直线解析式为．······················ 7分 .∴················ 9分．···························································· 10分 ∴当时.取得最大值.最大值为．······························································ 11分 (3)当取得最大值...∴．∴四边形是矩形．作点关于直线的对称点.连接．法一:过作轴于.交轴于点．设.则．在中.由勾股定理.．解得．∵.∴．由.可得.．∴． ∴坐标．································································································ 13分法二:连接.交于点.分别过点作的垂线.垂足为．易证．∴．设.则．∴.．由三角形中位线定理.． ∴.即． ∴坐标．······················································· 13分把坐标代入抛物线解析式.不成立.所以不在抛物线上．·················· 14分【查看更多】

题目列表(包括答案和解析)

阅读材料，然后解方程组．
材料：解方程组

x-y-1=0，①

4(x-y)-y=5．②

由①得x-y③，把③代入②，得4×1-y=5．
解得y=-1．
把y=-1代入③，得x=0．
∴

x=0

y=-1

这种方法称为“整体代入法”．你若留心观察，有很多方程组可采用此方法解答，请用这种方法解方程组

2x-3y-2=0，①

2x-3y+5

+2y=9．②

．