解:(1)∵直線y=kx+b與直線y=
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x平行,
∴k=
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,
由A的橫坐標為-4,得到A(-4,0),即OA=4,
將x=-4,y=0代入y=
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x+b得:
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×(-4)+b=0,
解得:b=2,
∴直線AB解析式為y=
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x+2,
令x=0,解得:y=2,
∴B(0,2),即OB=2,
在Rt△AOB中,根據(jù)勾股定理得:AB=
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=2
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,
又AD=
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,
則矩形ABCD的面積S=AD•AB=10;
(2)在Rt△ADH和Rt△BAO中,
∵∠HAD+∠ADH=90°,∠HAD+∠BAO=90°,
∴∠ADH=∠BAO,又∠DHA=∠BOA=90°,
∴△ADH∽△BAO,
∴
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=
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=
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,即
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=
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=
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=
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,
解得:DH=2,AH=1,
∴OH=OA+AH=4+1=5,
則D的坐標為(-5,2).
分析:(1)由直線y=kx+b與直線y=
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x平行,得出k的值,由A的橫坐標為-4,確定出A的坐標,將A的坐標代入直線AB解析式中求出b的值,確定出直線AB的解析式,令x=0,求出對應的y值,即為B的縱坐標,確定出OB的長,在直角三角形AOB中,利用勾股定理求出AB的長,再由AD的長,利用矩形的面積公式即可求出矩形ABCD的面積;
(2)由三角形ADH和三角形AOB中兩銳角互余,列出兩個等式,利用同角的余角相等得到一對角相等,再由一對直角相等,得出三角形ADH與三角形AOB相似,由相似得比例,將AD,AB,OA及OB的長代入,求出DH與AH的長,再由AH+OA求出OH的長,由D為第二象限點,即可得出D的坐標.
點評:此題屬于一次函數(shù)綜合題,涉及的知識有:待定系數(shù)法求一次函數(shù)解析式,一次函數(shù)與坐標軸的交點,坐標與圖形性質(zhì),相似三角形的判定與性質(zhì),勾股定理,以及矩形的面積公式,其中根據(jù)兩直線平行時k值相同得出k的值是本題的突破點.