Coa6 lottery forecast
∴OH=CB=3，
∴AH=OA-OH=6-3=3，
In Rt△ABH, BH=,
∴ The coordinates of point B are (3,6).
(2) as shown in figure 1, if the EG⊥x axis is at the G point, then EG//BH
∴△OEG∽△OBH
∴
OE = 2EB，
∴
∴
∴OG=2,EG=4，
The coordinate of point e is (2,4).
Similarly, the coordinate of point d is (0,5),
Let the analytical formula of straight line DE be y=kx+b,
rule
Solution, b=5
∴ The analytical formula of straight line DE is:
A: Yes.
① As shown in figure 1, when OD=DM=MN=NO=5, the quadrilateral ODMN is a diamond.
Make MP⊥y axis at point p,
MP//x axis,
∴△MPD∽△FOD，
∴
When y=0,
The solution is x= 10.
The coordinate of point ∴f is (10,0),
∴OF= 10。
At Rt△ODF,
∴
∴
The coordinate of point m is (-2,5+).
∴ The coordinate of point N is (-2,) ② As shown in Figure 2, when OD=DN=NM=MO=5, the quadrilateral ODNM is a diamond.
Extend the x axis of NM to point p, and then MP⊥x axis.
The point m is on the straight line y=- x+5,
∴ The coordinate of point m is (a, -a+5),
At Rt△OPM, OP 2 +PM 2 =OM 2.
∴a 2 +(- a+5) 2 =5 2, while a 1 =4, a 2 =0 (discarded),
∴ The coordinate of point m is (4,3),
The coordinate of point n is (4,8). ③ As shown in Figure 3, when OM=MD=DN=NO, the quadrilateral OMDN is a diamond.
Connect NM and OD at point p, and then divide NM and OD vertically.
∴y M =y N =OP=，
∴- x M +5=，
∴x M =5，
∴x N =-x M =-5，
The coordinate of point n is (-5,).
To sum up, there are three points n above the X axis, namely N 1 (-2,), n 2 (4,8) and n 3 (-5,).