OC=OA2+AC2=5,( 1)
∫AC∨OB,
∴△orq∽△crp(2 points)
∴OQPC=ORRC,
∴PC=32a,
∫OA+AC = 7, that is, 3a+32a=7,
∴ A = 149, (4 points)
AP=53, (5 points)
∴P point coordinates (53,3), Q point coordinates (149,0),
Let the functional relationship of the straight line PQ be y=kx+b,
∴ 53k+b = 3 149k+b = 0, and the solution is k = 27b =? Forty two.
Therefore, the functional relationship of the straight line PQ is y = 27x-42;; (8 points)
(2) When 0≤t≤ 157, point P is on OA and point Q is on OB.
S= 12×OQ×OP= 1225t2,
When 157≤t≤5, point P is on AC and point Q is on OB.
S = 12× OQ× BC = 65t, (4 points)
When 5 < t ≤ 70 1 1, points p and q are on BC.
S = 12× PQ× OB = 28-225t。 (6 points)