3, we can get AB = 2OB. ..
In Rt△AOB, OB= 12 and ab = 24 are obtained by Pythagorean theorem.
∴B(0, 12),
∫OA = 12
3,
∴A ( 12
3,0).
The analytical formula of straight line AB can be set as: y=ax+b,
∴b= 12 12
3a+b=0? ,
∴a=-
33b= 12? ,
∴ The analytical formula of available straight line AB is: y=-
33x+ 12。
(2) Connect the CD, if F is the FM⊥x axis of point M, then CB = CD.
∵∠OBA=90 -∠A=60,
△ CBD is an equilateral triangle.
∴BD=CB= 12OB=6,
∠BCD=60,∠OCD= 120。
∫ob is the diameter, OA⊥OB,
∴OA Cut C to O
⊙ dechuts ⊙ c to d,
Coe = CDE = 90, Organization of Eastern Caribbean States =.
∴∠OED=360 -∠COE-∠CDE-∠OCD=60。
∴∠OEC=∠DEC=30。
∴CE= 12,CO=6.
∴ in Rt△COE, Pythagorean Theorem OE=CE2-CO2=6.
3.
∵BG⊥EC in F,
∴∠GFE=90。
∠∠GBO+∠BGO =∠OEC+∠BGO,
∴∠GBO=∠OEC=30。
Therefore, FC= 12BC=3, EF=FC+CE= 15,
FM= 12EF= 152,ME=3FM= 15
32.
∴MO= 15
32-6
3=
three
32.
∴F(-
three
32, 152).
(3) Set the moving speed of Q point as VCM/s. 。
(i) When point P moves to the midpoint of AB, point Q moves to the midpoint of AO, PQ∑BC, PQ=BC, the quadrilateral CBPQ is a parallelogram, and point Q coincides with point E. 。
Available AP= 12, t=
AP4=3。
∴v=
AEt=
six
33=2
3 (cm/s).
(ii) When point P moves to BG midpoint and point Q moves to OG midpoint,
PQ∑BC, PQ=BC, quadrilateral CBPQ is a parallelogram.
Available OG=4
3,BG=8
3. Therefore PB=4.
3,OQ=2
3.
∴t=
AB+BP4=
24+4
34=6+
3.
∴v=
AQt=
12
3+2
36+
3=
28
3- 14 1 1 (cm/s).
The speed of point Q is 2.
3 cm/s or 28
3-1411cm/s comments: This topic mainly examines the comprehensive application of linear functions, the properties of parallelograms and the application of Pythagorean theorem, and classifies them according to the position of known point P motion.