(1)OH is the height on the hypotenuse of Rt△OPC, and ao = bo = √ 2;

= = & gt△AOB is isosceles Rt△,

= = & gtOH is the median line and vertical line on the hypotenuse of isosceles Rt△AOB.

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

P(-2) the straight line y = kx+b intersects the x axis at p (-2,0),

= = & gt0=-2k+b

= = & gtk = b/2；

Intersect with y axis at C. (0, oc)

= = & gtOC=-0*k+b

= = & gtb = OC

A and b are on a straight line, y = kx+b,

= => in Rt△PHO, OH= 1, OP=2,

= = & gt& ltHPO=30，

= => in Rt△PCO, 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

So:

1) Length of OH =1; k=3/√3，b = 2/√3；

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