在平面直角坐标系中,双曲线y=k/x(x>0,k≠0)的过A(6,1),.点B(a,b)(0<a<6)是双曲线上的一动点……

在平面直角坐标系中，如果双曲线y＝kx(k≠0)经过点（2，-1），那么k=______~

∵反比例函数y=kx（k≠0）经过点（2，-1），∴-1=k2，解得k=-2．故答案为：-2．

其实在算出点E的坐标后，还可以用另一种方法来算a²
∠ACE=∠ACB+∠FCE=90°，在菱形CDEF中,CE=√3CD
三角形ACE的面积=√3=1/2*AC*CE得出OC*CD=2从而得出a²=(√2+1)/2

（1）根据A点坐标求K=6，又C（0 6）
又据B在曲线上和BC=AC列两个方程，解a和b值
最后的D点设出来，又据AD=BC AB=CD解出来就可以了
（2）以AC做底，高就是b+1，那6*（b+1）=12
b=1这样也可以知道B（-6 -1）
继续求D点坐标 求最长的对角线即可

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这个题目并不难啊，y=k/x是反比例函数，又因为过点（6,1）所以y=6/x,你画个图就出来了，
A(6,1) B(a,b) C(0,1) D()把图画出来，我这里不是很方便，就没有继续，不难，努力1

完了不会

gte54yhfdxbctfr

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徐州市13683796402： 如图,在平面直角坐标系中,矩形OABC的顶点A、C分别在x轴和y轴上,点B的坐标为(6,4),双曲线y=kx(x>0)经过AB的中点D,且与BC交于点E,连接DE... - ？
亓秆生血：[答案] (1)∵点B的坐标为(6,4), ∴AB的中点D的坐标为(6,2), 将点D(6,2)的坐标代入y= k x(x>0), 得:k=6*2=12. ∵BC∥x轴, ∴点E的纵坐标与点B的纵坐标相等, ∴点E的纵坐标为4. ∵点E在双曲线上, ∴x= 12 4=3, ∴点E在坐标为(3,4). 设直线DE的...

徐州市13683796402： 如图所示,在平面直角坐标系中,双曲线y=kx(x<0)上有一点A(-2,2),AB⊥y轴于点B,点C是x轴正半轴上一动点,直线CB交双曲线于点D,DE⊥x轴于点E,连接... - ？
亓秆生血：[答案] (1)若四边形ADBE为菱形,则AB与DE互相垂直平分,由题意得,A(-2,2),B(0,2).则反比例函数的解析式是y=-4x,E(-1,0)D(-1,4).设直线BD的解析式是y=kx+b,将B(0,2),D(-1,4)代入y=kx+b,可得:2=...

徐州市13683796402： 如图,在平面直角坐标系中,双曲线y=k/x过第一象限内一点A,AB⊥x轴,垂足为B,S三角形AOB=2 - ？
亓秆生血： 1)设A(x,k/x),△AOB面积=(1/2)*x*(k/x)=k/2=2 解得k=42)由k=4,得直线y=x+4,当y=0时,x=-4 所以C(-4,0) 解方程组 y=4/x,y=x+4 解得x=-2±2√2,因为x>0 所以x=-2+2√2 y=2+2√2, 所以△ABC面积=(1/2)*(2+2√2)*4=4+4√23)因为当x所以y1

徐州市13683796402： 已知四边形ABCD是矩形,在平面直角坐标系中,点A,D的坐标分别是(0,2)(4,0)双曲线y=k/x经过长方形对角交点M,交AB与N,交BD于P请求出AN:NB与... - ？
亓秆生血：[答案] 因双曲线经过长方形对角线交点M,而M点坐标为(2,1) 所以双曲线为XY=2 故与AB的交点N为(1,2) 与BD的交点P为(4,1/2) 所以有AN=1 NB=3 DP=1/2 PB=3/2 所以AN:NB与DP:PB的比值1:3

徐州市13683796402： 如图,在平面直角坐标系中,直线y= - x - 4分别交x轴,y轴于A,B,交双曲线y=k/x(x<0)于M,连OM,且S△OBM=16 - ？
亓秆生血： 过程太长,而且有些东西比较麻烦,简单作个提示:第1问题:易知A(-4,0)、B(0,-4),所以OA=OB=4,由三角形OBM的面积是16,得MN=8,易知M(-8,4),从而K=-32.(同时可以看到:BO=ON=4,所以BA=AM=4√2.) 第2问题:作O关于直线...

徐州市13683796402： 如图,在平面直角坐标系中,双曲线y=k/x经过线段OA的中点C,过点A作AB⊥x轴于点B,交双曲线于点D,若△ACD的面积为6,则双曲线的解析式为? - ？
亓秆生血： 过点C作CE ⊥OB于点E,则S△OCE=1/2K,连接OD,则S△AOB=S△ACD+S△OCD+S△ODB=6+6+ 1/2K=12+1/2K,因为△OCE相似于△OAB,所以有1/2K:(12+1/2K)=1:4(即面积之比等于相似比的平方),所以解得K=8

徐州市13683796402： 如图,将一块直角三角板OAB放在平面直角坐标系中,B(2,0),∠AOB=60°,点A在第一象限,过点A的双曲线为y=k - ？
亓秆生血： 解:1)点O′与点A重合时,直线l垂直平分OA,如图,连PA,则PA=PO,∵d(2,0),∠AOd=60°,∴OB=2,∴AB= 3 OB=2 3 ,设u点坐标为(x,0),则uA=uO=x,uB=x-2,在Rt△PcB中,Pc2=PB2+uB2,即x2=(x-2)2+(2 3 )2,解得x=4,∴P点坐标为(4,0).2)4≤t≤2√5或-2√5 ≤t≤-4 赞同12|评论(11)

徐州市13683796402： 如图,在平面直角坐标系中.矩形ABCD的顶点A与坐标原点O重合.点B和点D分别在x轴与y轴的正半轴上.双曲线y= k x (x>0)经过顶点C,直线y= - x+2经过顶点B... - ？
亓秆生血：[答案] (1)∵B,E是直线y=-x+2与x轴,y轴的交点, ∴E(0,2),B(2,0), ∴OE=OB=2, ∵∠BOE=90°, ∴∠BEO=45°, 故答案为:45°; (2)在矩形ABCD中,AB=CD,AD=BC, ∵E(0,2),B(2,0), ∴OE=OB=2, ∴D(0,b),DE=EF=2-b, ∵∠1=∠2=45°, ∴△DEG是等腰直...

徐州市13683796402： 在平面直角坐标系中,双曲线y=k/x(x>0,k≠0)的过A(6,1),.点B(a,b)(0<a<6)是双曲线上的一动点,过点A作AC⊥y轴…… - ？
亓秆生血： 2:为菱形则AB=AC=6 作BP垂直AC,显然BP=a-1PA=6-(6/a) 则(a-1)^2+[6-(6/a)]^2=36 解这个二元一次方程就可以得出B,D就可以轻松求出 3:显然b=3 则B(2,3) 此时D(-4,3)或(8,2)若D为(-4,3) 易知CB此时=根号下2^2+2^2=2根号2,AD=根号下10^2+2^2=根号下104 若D为(8,3)则AB=根号下4^2+2^2=2根号5 CD=根号下8^2+2^2=根号下68 显然对角线最长为根号下104=2根号26

徐州市13683796402： 如图平行四边形ABcD的顶点A、B的坐标分别是A( - 1,0)B(0, - 2)顶点c.D在双曲线y=k/ - ？
亓秆生血： “四边形BCDE的面积是△ABE面积的5倍”,因同时有“四边形BCDE的面积是△ABD面积的2倍”,所以△ABD的面积是△ABE的面积的3倍,从而值 AD=3AE;∴ 在等腰直角△AGD中,AG=3AO,∴ OG=2AO=2*1=2,……D点横坐标是2,设纵坐标为y;由图看出,RT△CHD≌RT△AOB,∴ CH=AO=1,DH=BO=2,…C点横坐标是3,纵坐标y-2;∴ k=2*y=3*(y-2),解得 y=6,k=12;