Substitute the coordinates of two points: k = 34 and b = 0.
∴y=34x.
(2) When q moves on OC, let Q(m, 34m), that is, m2+(34m) 2 = (2t) 2, and the solution is m = 85t.
Then Q(85t, 65t) (0 ≤ t ≤ 5).
When Q moves on CB, the distance Q moves is 2t.
∫OC = 10，
∴CQ=2t- 10.
The abscissa of point ∴Q is 2t- 10+8 = 2t-2.
∴Q(2t-2,6)(5≤t≤ 10).
(3) The circumference of trapezoidal OABC is 44. When Q moves on OC, the distance of P is t and the distance of Q is (22-t).
△OPQ, the height on the OP surface is: (22? t)×35。
∴S△OPQ= 12t(22？ T)×35, S- trapezoid OABC =12 (18+10) × 6 = 84.
According to the meaning of the question: 12t(22? t)×35=84× 12。
Finishing: T2-22t+ 140 = 0.
∵△=222-4× 140