如图,在直角坐标系xoy中,△ABC的顶点坐标为A(—8,0),B(3,0),C(0,4),动点P从点A出发,沿着AB以每秒1个单
如图,在直角坐标系xoy中,△ABC的顶点坐标为A(—8,0),B(3,0),C(0,4),动点P从点A出发,沿着AB以每秒1个单
沿着AB以每秒1个单位长度的速度向终点B运动；动点Q从点B出发,沿着射线BC,以每秒1个单位长度的速度运动,当点P到达B时,点Q也停止运动,P,Q两点同时开始运动,设运动时间为t秒.
（1）当PQ⊥x轴,求此时t的值；
（2）设△BPQ的面积为S,求S关于t的函数关系式；
（3）当△APQ为等腰三角形时,求t的值；
（4）设△APQ的外接圆的圆心为M,当点C在圆M外时,写出t的取值范围.
(1)AB = 11,0 ≤ t ≤ 11t秒时,P(-8 + t,0),BQ = tOB = 3,OC = 4,BC = 5cos∠ABC = OB/BC = 3/5; sin∠ABC = OC/BC = 4/5Q的横坐标 = OB - BQcos∠ABC = 3 - 3t/5Q的纵坐标 = BQsin∠ABC = 4t/5当PQ⊥x轴时,P,Q的横...第3题的3是如何求出的，第4题能用初中知识吗AQ² = (3 - 3t/5 + 8)² + (4t/5 - 0)² = (11 - 3t/5)² + (4t/5)²PQ² = (-8 + t - 3 + 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)(iii)Q为顶点, AQ² = PQ²(11 - 3t/5)² + (4t/5)² = (8t/5 - 11)² + (4t/5)²(11 - 3t/5)²= (8t/5 - 11)²11 - 3t5 = 8t/5 - 11, 11t/5 = 22, t = 10或11 - 3t/5 = 11 - 8t/5, t = 0, 舍去