Question

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

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方向运动，点Q从点B出发，以每秒1个单位长度的速度沿x轴正方向运动。P、Q两点同时出发，设移动时间为t（t＞0）秒 （1）再点p，Q运动过程中，若△POQ与△AOB相似 求t的值

Answer 1

△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

有二解: t = 48/5或t = 25