Question

如图 在平面直角坐标系中 点o为坐标原点 直线Y=X+4交X轴于点A,交Y轴于点B,

Answer 1

(1)
A(-4,0),B(0,4),BC = 4,C(4,4)
y = - x + m过C,4 = -4 + m,m = 8
y = -x + 8,y = 0,x = 8,D(8,0)
折线C-B-O = 4 + 4 = 8,OD = 8,当点M到达点D时,点P同时到达点O
t秒时:OM = t,M(t,0),0 ≤ t ≤ 8
0 ≤ t ≤ 4:PC = t,P(4 - t,4)
4 P(0,8 - t)
OC的方程为y = x,0 ≤ t ≤ 4:Q(t,t)
4 (i) 容易看出,t = 4时,M(4,0),Q(4,4),P(0,4),此时OMQP为正方形（矩形的特例）.
(ii) t