$解：(2)②點(diǎn)B的“直角旋轉(zhuǎn)點(diǎn)”M 有兩個(gè)：$
$點(diǎn)P_{1}和P_{2}，作P_{1}H⊥x軸，P_{2}H⊥x軸，$
$垂足分別為H、H$
$∵點(diǎn)A、B的坐標(biāo)分別為(3，0)、(0，4)$
$∴OA=3，OB=4，∴AB= \sqrt{OA2+OB2}=5$
$∵點(diǎn)P_{1}是點(diǎn)B的“直角旋轉(zhuǎn)點(diǎn)”$
$∴AP=AP_{1}=5，∠BAP_{1}=90°$
$∴∠BAO+∠P_{1}AH_{1}=90°$
$∵∠BOA=90°，∴∠BAO+∠OBA=90°$
$∴∠OBA= ∠P_{1}AH$
$在△ABO 和△P_{1}AH 中$
$\begin{cases}{∠BOA=∠AH_{1}P_{1}}\\{∠OBA=∠P_{1}AH_{1}}\\{AB=P_{1}A}\end{cases}$
$∴△ABO≌△PA H(\mathrm {AAS})$
$∴AH_{1}=OB=4，P_{1}H_{1}=OA=3$
$∴點(diǎn)P_{1}的坐標(biāo)為(7，3)$
$同理可得點(diǎn)P_{2}的坐標(biāo)為(-1，-3)$
$設(shè)直線1的函數(shù)表達(dá)式為y=kx+b(k≠0)$
$將A(3，0)，P_{1}(7，3)代入，$
$得\begin{cases}{3k+b=0}\\{7k+b=3}\end{cases}，解得\begin{cases}{k=\frac {3}{4}}\\{b=\frac {9}{4}}\end{cases}$
$∴y=\frac{3}{4}x-\frac{9}{4}$
$∴點(diǎn)M的坐標(biāo)為(0，-\frac{9}{4})$
