(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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