如图1,已知数轴上有三点a,b,c.点a,c对应的数分别是-40和20,点b是ac的中点.(1)

Question

如图,已知数轴上有三点A、B、C,它们对应的数分别为a、b、c,且c-b=b-a；点C对应的数是20.
（1）若BC=30,求a、b的值；
（2）在（1）的条件下,动点P,Q分别从AC两点左运动,同时动点R从A点出发向右运动,点P、R、Q的速度分别为8个单位长度/秒、4个单位长度/秒、2个单位长度/秒,点M为线段PR的中点,点N为线段RQ的中点,在R、Q相遇前,多少秒时恰好满足MR=4RN；
（3）在（1）的条件下,O为原点,动点P、Q分别从A、C同时运动,P向左运动,Q向右运动,P点的运动速度为8个单位长度/秒,点Q的速度为4个单位长度/秒,N为OP的中点,M为BQ的中点,在P、Q的运动过程中,PQ与MN的长存在一个确定的相等关系,请指出他们的关系,并说明理由.

Answer 1

1)
c - b = b - a, B为AC的中点, BC = 30 = 20 - b, b = -10
b - a = -10 -a = 30, a = -40
(2)
t秒时bai,
p = -40 - 8t
r = -40 + 4t
q = 20 - 2t
m = (-40 - 8t - 40 + 4t)/2 = -40 -2t
n = (-40 + 4t + 20 - 2t)/2 = -10 + t
MR = (-40 + 4t) - (-40 -2t) = 6t
RN = (-10 + t) - (-40 + 4t) = 30 - 3t
MR = 4RN: 6t = 4(30 - 3t), t = 20/3
(3)
t秒时,
p = -40 - 8t
q = 20 + 4t
n = (0 - 40 - 8t)/2 = -20 - 4t
m = (-10 + 20 + 4t)/2 = 5 + 2t
PQ = 20 + 4t - (-40 - 8t) = 60 + 12t = 12(5 + t)
MN = (5 + 2t) - (-20 - 4t) = 25 + 6t本回答被网友采纳

Answer 2

(1)
c - b = b - a, B为AC的中点, BC = 30 = 20 - b, b = -10
b - a = -10 -a = 30, a = -40
(2)
t秒时,
p = -40 - 8t
r = -40 + 4t
q = 20 - 2t
m = (-40 - 8t - 40 + 4t)/2 = -40 -2t
n = (-40 + 4t + 20 - 2t)/2 = -10 + t
MR = (-40 + 4t) - (-40 -2t) = 6t
RN = (-10 + t) - (-40 + 4t) = 30 - 3t
MR = 4RN: 6t = 4(30 - 3t), t = 20/3
(3)
t秒时,
p = -40 - 8t
q = 20 + 4t
n = (0 - 40 - 8t)/2 = -20 - 4t
m = (-10 + 20 + 4t)/2 = 5 + 2t
PQ = 20 + 4t - (-40 - 8t) = 60 + 12t = 12(5 + t)
MN = (5 + 2t) - (-20 - 4t) = 25 + 6t
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