|xa-200|=2|xa-xb| |xb-200|=300, then xb=500 or-100 is because XB < 200 is xb=- 100.

Then xa =-400,0 is due to xa.

2.P(-400) Q(200)R(-400) is the initial coordinate, and when the time is t, the coordinate is t seconds.

p(-400- 10t)Q(200-5t)R(-400+2t)

Then m (-400-4t) n (-100-1.5t).

Then MR=4RN has |6t|=|3.5t-300|.

Because the situation after point R meets point Q is not considered, -400+2t.

Simplified to 6t=300-3.5t, t=300/9.5.

3, E(-800)D(0) initial P(-800) Q(0) time is t.

Then p (-800-10t) q (-5t) m (-400-7.5t).

3/2QC-AM = 1.5 * | 200+5t |-| 7.5t |

In the process of moving from point D to point A, point Q is -400.

It can be seen that 3/2QC-AM=300+7.5t-7.5t=300.