$ 解:(1)由题,2a+8=0,解得a=-4,∴P(-6,0)$ $(2)由题,a-2=1,解得a=3 ∴P(1,14)$ $(3)由题,|a-2|=|2a+8|,解得a=-2或-10$ $当a=-2时,P(-4,4)$ $当a=-10时,P(-12,-12)$ $综上,P(-4,4)或(-12,-12)$
$ 解:(2)由题\begin{cases}{ 5x+y=-9 }\ \\ { x+5y=3 } \end{cases}解得\begin{cases}{ x=-2 }\ \\ { y=1 } \end{cases}∴P(-2,1)$ $(3)由题,P_{1}(c-1,2c)$ $则P_{2}(3-c,-5c-1)$ $由题,3-c=0或-5c-1=0,解得c=3或-\frac {1}{5}$ $当c=3时,P_{2}(0,-16)$ $当c=-\frac {1}{5}时,P_{2}(\frac {16}{5},0)$ $综上,P_{2}(0,-16)或(\frac {16}{5},0)$
$解:(1)设t秒后PQ//y轴$ $此时不难得出,AP=OQ,即9-2t=t,解得t=3$ $∴3秒后PQ//y轴$ $(2)当P在y轴右侧时,$ $S_{四边形AOQP}=\frac {1}{2}(AP+OQ)×AO=10$ $即2(9-t)=10,解得t=4\ \ ,此时AP=9-2t=1\ ,P(1,4)$ $当P在y轴左侧时,AP=2t-9,同理有\frac {1}{2}(AP+OQ)×AO=10$ $即2(3t-9)=10,\ \ 解得t=\frac {14}{3}\ ,此时AP=2t-9=\frac {1}{3},\ \ P(-\frac {1}{3},4)$ $综上,P(1,4)或(-\frac {1}{3},4)$
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