答案：10.$\dfrac{15}{2}$

答案：11.解:设经过$(1,4),(2,7)$两点的直线的函数表达式为$y=kx+b$,$\therefore \begin{cases} k+b=4,\\ 2k+b=7, \end{cases}$解得$\begin{cases} k=3,\\ b=1, \end{cases}$$\therefore y=3x+1$,
将$(a,10)$代入函数表达式,得$3a+1=10$,解得$a=3$.

答案：
12.解:(1)如所示.
(2)如答图,$\because$点$C$与点$C'$关于$x$轴对称,$\therefore C'(4,-2)$.
$\because B(0,4)$,$\therefore$可设直线$BC'$的函数表达式为$y=kx+4$,
则$4k+4=-2$,解得$k=-1.5$,
$\therefore$直线$BC'$的函数表达式为$y=-1.5x+4$.

答案：
13.(1)$(8,0)$ $(0,-6)$
(2)解:设$OC$长为$m$,则$CA=OA-OC=8-m$,
$\because BC$平分$∠ OBA$,$CD⊥ AB$,$\therefore CD=OC=m$,
$\therefore$在$\mathrm{Rt}△ COB$和$\mathrm{Rt}△ CDB$中,$\begin{cases} CO=CD,\\ BC=BC, \end{cases}$
$\therefore \mathrm{Rt}△ COB≌\mathrm{Rt}△ CDB(\mathrm{HL})$,$\therefore BD=OB=6$.
在$\mathrm{Rt}△ AOB$中,
由勾股定理,得$AB=\sqrt{OA^2+OB^2}=10$,
$\therefore AD=AB-BD=4$.
$\therefore$在$\mathrm{Rt}△ ACD$中,$CD^2+AD^2=AC^2$,
即$m^2+4^2=(8-m)^2$,
解得$m=3$,$\therefore OC=CD=3$,$AC=5$.
$\because S_{△ ACB}=\frac{1}{2}AC· OB=15$,$S_{△ BCD}=\frac{1}{2}BD· CD=9$,
$\therefore S_{△ ACD}=S_{△ ACB}-S_{△ BCD}=6$,
即$\frac{1}{2}AC· |y_D|=\frac{1}{2}×5×|y_D|=6$,解得$|y_D|=\frac{12}{5}$,
$\therefore y_D=-\frac{12}{5}$,
将$y=-\frac{12}{5}$代入$y=\frac{3}{4}x-6$,得$-\frac{12}{5}=\frac{3}{4}x-6$,
解得$x=\frac{24}{5}$,$\therefore$点$D$的坐标为$(\frac{24}{5},-\frac{12}{5})$.
(3)解:如,连接$DF$,$\because BC$平分$∠ OBA$,$CD⊥ AB$,
$\therefore$点$O,D$关于$BC$对称,$\therefore OF=DF$,
$\therefore EF+OF=EF+DF$,即点$D$到$y$轴的距离为$EF+OF$最小值,$\therefore EF+OF$的最小值为$\frac{24}{5}$.