题109在直角坐标系中,定义两点P(x_1,y_1)与Q(x_2,y_2)之间的“直角距离”为d(P,Q)=|x_1-x_2|+|y_1-y_2|.现给出四个命题:(1)已知P(1,3),Q(sin~2α,cos~2α)(α∈R),则d(P,Q)为定值;(2)已知P,Q,R三点不共线,则必有d(P,Q)+d(Q,R)>d(P,R);(3)用|PQ|表示P,Q两点间的距离,那么|PQ|≥2~(1/2)/2d(P,Q); Problem 109 In the Cartesian coordinate system, we define the distance between the two points P (x_1, y_1) and Q (x_2, y_2) as d (P, Q) = | x_1-x_2 | + | y_1- y 2 |. Now four propositions are given: (1) d (P, Q) is fixed if P (1, 3) and Q (sin ~ 2α, cos ~ 2α) 2) Knowing that P, Q and R are not collinear at all three points, then we must have d (P, Q) + d (Q, R)> d PQ | ≥2 ~ (1/2) / 2d (P, Q);