2008年苏州数学试题

(1)OH是Rt△OPC的斜边上的高度，ao = bo =√2；

= = & gt△AOB是等腰Rt△，

= = & gtOH是等腰Rt△AOB的斜边上的中线和垂直线。

= = & gtAH = HB = = OH = AO/√2 =√2/√2 = 1；

P(-2)直线y = kx+b与x轴相交于p (-2，0)，

= = & gt0=-2k+b

= = & gtk = b/2；

在C. (0，oc)处与y轴相交

= = & gtOC=-0*k+b

= = & gtb = OC

a和b在直线上y = kx+b，

= = & gt在Rt△PHO中，OH=1，OP=2，

= = & gt& ltHPO=30，

= = & gt在Rt△PCO中，op = 2，

= = & gtOC/OP=1/√3，

= = & gtOC = OP * 1/√3 = 2 * 1/√3 = 2/√3，

= = & gtb=OC=2/√3

= = & gtk=b/2=(2/√3)/2=3/√3

所以:

1)OH的长度= 1；k=3/√3，b = 2/√3；

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