如图,在直角坐标系平面中,O为原点,A(0,6),B(8,0).点p从点A出发,以每秒2个单位长度的速度沿射线AO方向运

△POQ与△AOB均为直角三角形,tanOBA = OA/OB = 6/8 = 3/4
若△POQ与△AOB相似,只需tanOQP = 3/4或tanOPQ = 3/4即可.
t时,P(0,6 - 2t),Q(8 + t,0)
(1) 0 < t < 3
P在OA上,OP = 6 - 2t,OQ = 8+t
tanOQP = OP/OQ = (6-2t)/(8+t) = 3/4,t= 0,舍去
或tanOPQ = OQ/OP = (8+t)/(6-2t) = 3/4,t = -7/5 < 0,舍去
(2) t > 3
P在x轴下方,OP = 2t - 6,OQ = 8+t
tanOQP = OP/OQ = (2t - 6)/(8+t) = 3/4,t = 48/5
或tanOPQ = OQ/OP = (8+t)/(2t - 6) = 3/4,t =25
t = 48/5或t = 25