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第174頁(yè)

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$y = 3x - 6$

$解:(2)①將x=0代入y=3x?6,得y=?6$

$∴E(0,?6)$

$BQ將△BDE的分為1:2兩部分時(shí):$

$當(dāng)S_{△BEQ}:S_{△BDQ}=1:2時(shí)$

$S_{△BQE}:S_{△BDE}=1:3$

$∴EQ=\frac{1}{3}DE$

$∵D(4,6),E(0,?6),∴點(diǎn)Q的橫坐標(biāo)為0+\frac{1}{3}×(4?0)=\frac{4}{3}$

$縱坐標(biāo)為?6+\frac{1}{3}×(6+6)=?2,∴Q(\frac{4}{3}?2)$

$當(dāng)S_{△BDQ}:S_{△BEQ}=1:2時(shí),S_{△BQE}:S_{△BDE}=2:3$

$∴EQ=\frac{2}{3}DE$

$∵D(4,6),E(0,?6),∴點(diǎn)Q的橫坐標(biāo)為0+\frac{2}{3}×(4?0)=\frac{8}{3}$

$縱坐標(biāo)為?6+3×(6+6)=2,∴Q(\frac{8}{3},2)$

$綜上可知,點(diǎn)Q的坐標(biāo)為(\frac{4}{3}?2)或(\frac{8}{3},2)$

$②(3,3)或(\frac {18}{7},\frac {12}{7})$

$解:(1)在y=\frac{3}{2}x+3中,令x=0得y=3,令y=0得x=?2$

$∴A(?2,0), B(0,3)$

$∵點(diǎn)C是點(diǎn)A關(guān)于y軸的對(duì)稱點(diǎn)，C(2,0)$

$把D(?1,a)代入y=\frac{3}{2}x+3,得a=?\frac{3}{2}+3=\frac{3}{2}$

$∴D(?1,\frac{3}{2})$

$∴可求得直線CD的表達(dá)式為y=-\frac {1}{2}x+1$

$(2)存在$

$設(shè)AP交y軸于點(diǎn)T,設(shè)點(diǎn)P(m,-\frac{1}{2}m+1)$

$同理,由點(diǎn)A,P的坐標(biāo)得直線AP的表達(dá)式為y=\frac {-\frac {1}{2}m+1}{m+2}(x+2)$

$則點(diǎn)T(0,\frac {-m+2}{m+2}),BT=3+\frac {m-2}{m+2}$

$∴S_{△PAB}=\frac {1}{2}×BT×|x_{P}-x_{A}|=4S_{△COD}$

$=4×\frac {1}{2}×2×\frac {3}{2},解得m=2或-4$

$∴P(2,0)或(-4,3)$

$(3)(-\frac {22}{5},\frac {16}{5})或(\frac {190}{31},-\frac {64}{31})$

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