Question

如图，在平面直角坐标系中，直线y=—3/4x+6分别交x轴、y轴于C,A两点。将射线AM绕着点A顺时针旋转45°得到

射线AN。点D为AM上的动点，点B为AN上的动点，点C在∠MAN的内部。 （1）求线段AC的长 （2）当AM平行x轴，且四边形ABCD为等腰梯形时，求D的坐标(3)若再图1中存在点B,D，使三角形ABD周长最小，则此最小值为多少 （知道答案是 根号200 求完整过程 谢谢！）

Answer 1

1)y=-3/4+6
A的坐标为（0，6），C的坐标为（0，8）因为角AOC=90° ，所以AC=根号下6方+8方=10
（2)取AD，BC的中点，E,F，连结EF.
因为AM//OC,所以角ABO=角DAB=45°，因为角AOC=90°，所以AO=OB=6,则BC=8-6=2.因为ADCB为等腰梯形，所以A,D关于EF对称，因为OB=6,BC=2,所以OF=6+2/2=7,则D的横坐标...

Answer 2

作点C关于AM对称点E,作点C关于AN对称点F,连接EF交AM于D,交AN于B,连接CB、CD,则三角形CBD为周长最小，此时周长为EF长。由对称性知AE=AC=10，AF=AC=10，∠EAF==2∠MAN=90°,由勾股定理得最小值为10根号2

Answer 3

英语词组有两个的抄一个？？

Answer 4

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Answer 5

图呢，没图没真相