Simone如图，一次函数$y=kx+b\left(k\neq 0\right)$与反比例函数$y=\frac{a}{x}({a≠0})$的图象在第一象限交于$A$、$B$两点，$A$点的坐标为$\left(m,4\right)$，$B$点的坐标为$\left(3,2\right)$，连接$OA$、$OB$，过$B$作$BD\bot y$轴，垂足为$D$，交$OA$于$C$.若$OC=CA$，
的有关信息介绍如下：
$\left(1\right)\because $点$B\left(3,2\right)$在反比例函数$y=\frac{a}{x}$的图象上，$\therefore a=3\times 2=6$，$\therefore $反比例函数的表达式为$y=\frac{6}{x}$，$\because $点$A$的纵坐标为$4$，$\because $点$A$在反比例函数$y=\frac{6}{x}$图象上，$\therefore A(\frac{3}{2}$，$4)$，$\therefore \left\{\begin{array}{l}{3k+b=2}\\{\frac{3}{2}k+b=4}\end{array}\right.$，$\therefore \left\{\begin{array}{l}{k=-\frac{4}{3}}\\{b=6}\end{array}\right.$，$\therefore $一次函数的表达式为$y=-\frac{4}{3}x+6$；$(2)$如图$1$，过点$A$作$AF\bot x$轴于$F$交$OB$于$G$，$\because B\left(3,2\right)$，$\therefore $直线$OB$的解析式为$y=\frac{2}{3}x$，$\therefore G(\frac{3}{2}$，$1)$，$A(\frac{3}{2}$，$4)$，$\therefore AG=4-1=3$，$\therefore S_{\triangle AOB}=S_{\triangle AOG}+S_{\triangle ABG}=\frac{1}{2}\times 3\times 3=\frac{9}{2}$.$(3)$如图$2$中，①当$\angle AOE_{1}=90^{\circ}$时，$\because $点$A(\frac{3}{2}$，$4)$，$\therefore $直线$AC$的解析式为$y=\frac{8}{3}x$，$\therefore $直线$OE_{1}$的解析式为$y=-\frac{3}{8}x$，当$y=2$时，$x=-\frac{16}{3}$，$\therefore E_{1}(-\frac{16}{3}$，$2)$；②当$\angle OAE_{2}=90^{\circ}$时，可得直线$AE_{2}$的解析式为$y=-\frac{3}{8}x+\frac{73}{16}$，当$y=2$时，$x=\frac{41}{6}$，$\therefore E_{2}(\frac{41}{6}$，$2)$.综上所述，满足条件的点$E$坐标为$(-\frac{16}{3}$，$2)$或（$\frac{41}{6}$，$2)$.
$\left(1\right)\because $点$B\left(3,2\right)$在反比例函数$y=\frac{a}{x}$的图象上，$\therefore a=3\times 2=6$，$\therefore $反比例函数的表达式为$y=\frac{6}{x}$，$\because $点$A$的纵坐标为$4$，$\because $点$A$在反比例函数$y=\frac{6}{x}$图象上，$\therefore A(\frac{3}{2}$，$4)$，$\therefore \left\{\begin{array}{l}{3k+b=2}\\{\frac{3}{2}k+b=4}\end{array}\right.$，$\therefore \left\{\begin{array}{l}{k=-\frac{4}{3}}\\{b=6}\end{array}\right.$，$\therefore $一次函数的表达式为$y=-\frac{4}{3}x+6$；$(2)$如图$1$，过点$A$作$AF\bot x$轴于$F$交$OB$于$G$，$\because B\left(3,2\right)$，$\therefore $直线$OB$的解析式为$y=\frac{2}{3}x$，$\therefore G(\frac{3}{2}$，$1)$，$A(\frac{3}{2}$，$4)$，$\therefore AG=4-1=3$，$\therefore S_{\triangle AOB}=S_{\triangle AOG}+S_{\triangle ABG}=\frac{1}{2}\times 3\times 3=\frac{9}{2}$.$(3)$如图$2$中，①当$\angle AOE_{1}=90^{\circ}$时，$\because $点$A(\frac{3}{2}$，$4)$，$\therefore $直线$AC$的解析式为$y=\frac{8}{3}x$，$\therefore $直线$OE_{1}$的解析式为$y=-\frac{3}{8}x$，当$y=2$时，$x=-\frac{16}{3}$，$\therefore E_{1}(-\frac{16}{3}$，$2)$；②当$\angle OAE_{2}=90^{\circ}$时，可得直线$AE_{2}$的解析式为$y=-\frac{3}{8}x+\frac{73}{16}$，当$y=2$时，$x=\frac{41}{6}$，$\therefore E_{2}(\frac{41}{6}$，$2)$.综上所述，满足条件的点$E$坐标为$(-\frac{16}{3}$，$2)$或（$\frac{41}{6}$，$2)$.
$\left(1\right)\because $点$B\left(3,2\right)$在反比例函数$y=\frac{a}{x}$的图象上，$\therefore a=3\times 2=6$，$\therefore $反比例函数的表达式为$y=\frac{6}{x}$，$\because $点$A$的纵坐标为$4$，$\because $点$A$在反比例函数$y=\frac{6}{x}$图象上，$\therefore A(\frac{3}{2}$，$4)$，$\therefore \left\{\begin{array}{l}{3k+b=2}\\{\frac{3}{2}k+b=4}\end{array}\right.$，$\therefore \left\{\begin{array}{l}{k=-\frac{4}{3}}\\{b=6}\end{array}\right.$，$\therefore $一次函数的表达式为$y=-\frac{4}{3}x+6$；$(2)$如图$1$，过点$A$作$AF\bot x$轴于$F$交$OB$于$G$，$\because B\left(3,2\right)$，$\therefore $直线$OB$的解析式为$y=\frac{2}{3}x$，$\therefore G(\frac{3}{2}$，$1)$，$A(\frac{3}{2}$，$4)$，$\therefore AG=4-1=3$，$\therefore S_{\triangle AOB}=S_{\triangle AOG}+S_{\triangle ABG}=\frac{1}{2}\times 3\times 3=\frac{9}{2}$.$(3)$如图$2$中，①当$\angle AOE_{1}=90^{\circ}$时，$\because $点$A(\frac{3}{2}$，$4)$，$\therefore $直线$AC$的解析式为$y=\frac{8}{3}x$，$\therefore $直线$OE_{1}$的解析式为$y=-\frac{3}{8}x$，当$y=2$时，$x=-\frac{16}{3}$，$\therefore E_{1}(-\frac{16}{3}$，$2)$；②当$\angle OAE_{2}=90^{\circ}$时，可得直线$AE_{2}$的解析式为$y=-\frac{3}{8}x+\frac{73}{16}$，当$y=2$时，$x=\frac{41}{6}$，$\therefore E_{2}(\frac{41}{6}$，$2)$.综上所述，满足条件的点$E$坐标为$(-\frac{16}{3}$，$2)$或（$\frac{41}{6}$，$2)$.
