In the plane rectangular coordinate system, point a (0,4) B (3,4) C (6,0), moving point P starts from point a and moves downward on the y-axis at a speed of 1 unit / second Starting from point C, the moving point Q moves to the left on the x-axis at a speed of 2 units per second, passing through point P as RP, perpendicular to the y-axis, intersecting ob with R, connecting RQ. When point P coincides with point O, both moving points stop moving. Let the time of motion be T seconds. (1) if t = 1, find the coordinates of point R; (2) if 3

In the plane rectangular coordinate system, point a (0,4) B (3,4) C (6,0), moving point P starts from point a and moves downward on the y-axis at a speed of 1 unit / second Starting from point C, the moving point Q moves to the left on the x-axis at a speed of 2 units per second, passing through point P as RP, perpendicular to the y-axis, intersecting ob with R, connecting RQ. When point P coincides with point O, both moving points stop moving. Let the time of motion be T seconds. (1) if t = 1, find the coordinates of point R; (2) if 3

（1）∵A（0,4）,B（3,4）,
⊥ ab ⊥ Y axis, ab = 3
∵ RP ⊥ Y axis,
∴∠OPR=∠OAB=90°．
And ∠ por = ∠ AOB,
∴△OPR∽△OAB,
∴$\frac{OP}{OA}=\frac{PR}{AB}$．
When t = 1, AP = 1, Op = 3,
∴$\frac{3}{4}=\frac{PR}{3}$,
∴$PR=\frac{9}{4}$．
The ordinate of R is equal to the length of Op,
The coordinates of point R are ($\ frac {9} {4} $, 3)
(2) As shown in the figure, if the BD ⊥ X axis passes through point B at point D, then d (3,0)
In △ BOC,
∵ od = DC = 3, and BD ⊥ OC,
∴OB=BC．
∵△OPR∽△OAB,
∴$\frac{OR}{OB}=\frac{OP}{OA}$,
∵ in RT △ OBD, $ob = - sqrt {o {d ^ 2} + B {d ^ 2} = 5$
∴$\frac{OR}{5}=\frac{4-t}{4}$,
∴$OR=\frac{20-5t}{4}$．
AP = t, CQ = 2T (0 ≤ t ≤ 4)
The discussion is divided into three situations
① When 0 ≤ t ＜ 3, i.e. point Q moves from point C to point O (not coincident with O),
∵OB=BC
∴∠BOC=∠BCO＞∠BCA
∵ ab ∥ X axis,
∴∠BOC=∠ABO,∠BAC=∠ACO,
∵∠ABO＜ABC,∠BCO＞∠ACO,
∴∠BOC＜ABC,∠BOC＞∠BAC,
When 0 ≤ t ＜ 3, △ ORQ and △ ABC cannot be similar
② When t = 3 and point Q coincides with O, △ ORQ becomes line or, so it is impossible to be similar to △ ABC
③ As shown in the figure, when 3 ＜ t ≤ 4, that is, when the point Q moves to the left from the origin o,
∵ BD ∥ Y axis
∴∠AOB=∠OBD
∵OB=BC,BD⊥OC
∴∠OBD=∠DBC
∴∠QOR=90°+∠AOB=90°+∠DBC=∠ABC9
When $- frac {OQ} {or} = - frac {AB} {BC} $,
∵OQ=2t-6,
∴$\frac{2t-6}{{\frac{20-5t}{4}}}=\frac{3}{5}$,
∴$t=\frac{36}{11}$．
When $- frac {OQ} {or} = - frac {BC} {AB} $,
Similarly, we can get $t = - frac {172} {49} $
The results show that both $t = - frac {36} {11} and $t = - frac {172} {49} are within 3 ＜ t ≤ 4,
The values of all t that meet the requirements are $- frac {36} {11} $and $- frac {172} {49} $

As shown in the figure, in the plane rectangular coordinate system, point a (0,2) is known, and point P is a moving point on the x-axis. Take the line AP as one side, and make an equilateral triangle Apq on one side. When point P moves to the origin o, the position of Q is recorded as B. (1) calculate the coordinates of point B; (2) verify that when point P moves on the x-axis (P does not coincide with Q), ABQ is the fixed value; (3) whether there is a point P, Make the quadrilateral with a, O, Q and B as vertex trapezoid? If it exists, ask for the coordinates of point p; if it does not exist, explain the reason

(1) Passing through point B as BC ⊥ Y axis at point C,
∵ a (0,2), △ AOB are equilateral triangles,
∴AB=OB=2,∠BAO=60°,
Ψ BC = root 3, OC = AC = 1,
B (root 3,1);
(2) When point P moves on the x-axis (P does not coincide with O), it does not lose generality,
∵∠PAQ═∠OAB=60°,
∴∠PAO=∠QAB,
In △ apo and △ AQB,
∵AP=AQ,∠PAO=∠QAB,AO=AB
The total establishment of ≌ apo ≌ AQB,
The results show that ﹥ ABQ = ﹥ AOP = 90 ° is always true,
When point P moves on the x-axis (P does not coincide with Q), ABQ is the fixed value of 90 °;
(3) It can be seen from (2) that point q is always on the straight line passing through point B and perpendicular to AB, so AO and BQ are not parallel
① When p is on the negative half axis of X, q is below B,
In this case, if ab ‖ OQ, the quadrilateral aoqb is a trapezoid,
When ab ∥ OQ, bqo = 90 ° and bqo = ABO = 60 °
If ob = OA = 2, BQ = root 3 can be obtained,
It can be seen from (2) that △ apo ≌ △ AQB,
Ψ OP = BQ = root 3,
The coordinate of P is (- root 3,0)
② When p is on the positive half axis of X, q is above B,
In this case, if AQ ‖ ob, the quadrilateral aoqb is a trapezoid,
When AQ ‖ ob, ∠ ABQ = 90 ° and ∠ qAB = ∠ ABO = 60 °
If AB = 2, BQ = 2, 3,
It can be seen from (2) that △ apo ≌ △ AQB,
Ψ OP = BQ = 2, 3,
The coordinates of P are (2 3,0)
In conclusion, the coordinates of P are (- root 3,0) or (root 3,0)

In the plane rectangular coordinate system, it is known that a (- 1,2), B (2,1) find a point P on the x-axis to minimize the value of PA + PA, and calculate the coordinates of point P

Let A1 (- 1, - 2),
For any point P on the x-axis, since a and A1 are symmetric about the x-axis, PA = PA1
Therefore, PA + Pb = PA1 + Pb > = A1B = √ [(2 + 1) ^ 2 + (1 + 2) ^ 2] = 3 √ 2
If and only if A1, P, B are in a straight line, PA + Pb is the minimum
Let P (x, 0), then
(0+2)/(x+1)=(0-1)/(x-2)
The solution is x = 1, so p (1,0)

As shown in the figure, in the plane rectangular coordinate system, the coordinates of point a are (0, 2), point P (T, 0) is on the x-axis, and B is the midpoint of line Pa. rotate line Pb clockwise about point P to get line PC, connecting OB and BC (1) (2) when t > 0, can a quadrilateral with vertices P, O, B and C be a parallelogram? If you can, find the corresponding value of T? If not, please explain the reason; (3) when t is what value, △ AOP is similar to △ APC?

(1) When ob ⊥ BP, the quadrilateral with vertex P, O, B and C is flat