$ 解:(1)由題,2a+8=0,解得a=-4,∴P(-6,0)$ $(2)由題,a-2=1,解得a=3 ∴P(1,14)$ $(3)由題,|a-2|=|2a+8|,解得a=-2或-10$ $當(dāng)a=-2時(shí),P(-4,4)$ $當(dāng)a=-10時(shí),P(-12,-12)$ $綜上,P(-4,4)或(-12,-12)$
$ 解:(2)由題\begin{cases}{ 5x+y=-9 }\ \\ { x+5y=3 } \end{cases}解得\begin{cases}{ x=-2 }\ \\ { y=1 } \end{cases}∴P(-2,1)$ $(3)由題,P_{1}(c-1,2c)$ $則P_{2}(3-c,-5c-1)$ $由題,3-c=0或-5c-1=0,解得c=3或-\frac {1}{5}$ $當(dāng)c=3時(shí),P_{2}(0,-16)$ $當(dāng)c=-\frac {1}{5}時(shí),P_{2}(\frac {16}{5},0)$ $綜上,P_{2}(0,-16)或(\frac {16}{5},0)$
$解:(1)設(shè)t秒后PQ//y軸$ $此時(shí)不難得出,AP=OQ,即9-2t=t,解得t=3$ $∴3秒后PQ//y軸$ $(2)當(dāng)P在y軸右側(cè)時(shí),$ $S_{四邊形AOQP}=\frac {1}{2}(AP+OQ)×AO=10$ $即2(9-t)=10,解得t=4\ \ ,此時(shí)AP=9-2t=1\ ,P(1,4)$ $當(dāng)P在y軸左側(cè)時(shí),AP=2t-9,同理有\(zhòng)frac {1}{2}(AP+OQ)×AO=10$ $即2(3t-9)=10,\ \ 解得t=\frac {14}{3}\ ,此時(shí)AP=2t-9=\frac {1}{3},\ \ P(-\frac {1}{3},4)$ $綜上,P(1,4)或(-\frac {1}{3},4)$
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