第109页 - 创新优化学案八年级数学苏科版

112

$ 解:(1)易求A(4,0),B(0,3)$

$把A坐标代入y=kx-\frac {16}{3}得\ 4k-\frac {16}{3}=0,解得k=\frac {4}{3}$

$∴y=\frac {4}{3}x-\frac {16}{3},可求得C(0,-\frac {16}{3})$

$(2)可设P(a,-\frac {3}{4}a+3)\ (0\lt a\lt 4)$

$由题有,\frac {1}{2}×(3+\frac {16}{3})×(4-a)=10$

$解得a=\frac {8}{5}$

$∴P(\frac {8}{5},\frac {9}{5})$

$(3)（更多请点击查看作业精灵详解）$

$解:如图①$

$当△AMN≌△AOC时,AO=AM=4$

$∴M(8,0)$

$C,N关于A点对称$

$∴N(8,\frac {16}{3})$

$如图②$

$当△ANM≌△AOC时$

$AM=AC=\sqrt {OC^{2}+OA^{2}}=\frac {20}{3}$

$设N(t,\frac {4}{3}t-\frac {16}{3})$

$则AN=AO=4=\sqrt {(t-4)^{2}+(\frac {4}{3}t-\frac {16}{3})^{2}}$

$解得t=\frac {32}{5},\frac {4}{3}t-\frac {16}{3}=\frac {16}{5}$

$∴N(\frac {32}{5},\frac {16}{5})$

$如图③$

$此种情况与情况②类似,即这两个△AMN$

$关于点A中心对称$

$∴可求得N(\frac {8}{5},-\frac {16}{5})$

$综上,N(8,\frac {16}{3})或(\frac {32}{5},\frac {16}{5})或(\frac {8}{5},-\frac {16}{5})$