The points A (1, 2), B(3, -4) are two vertices of the rectangle ABCD. The point P (3, 8) lies on CD produced, then C=

$\begin{array}{l}Givenpo\mathrm{int}sA(1,2)B(3,-4)andP(3,8)\\ slpoeofAB=\frac{-4-2}{3-1}=\frac{-6}{2}=-3\\ BC\perp AB\\ Equationof\overline{BC}isy+4=\frac{-1}{-3}(x-3)\\ \Rightarrow 3y+12=x-3\\ \Rightarrow x-3y-15=0--\left(1\right)\\ \overline{CD}\parallel \overline{AB}\\ \therefore Equationof\overline{CD}isy-8=-3(x-3)\\ \Rightarrow y-8=-3x+9\\ \Rightarrow 3x+y-17=0--\left(2\right)\end{array}$

$\begin{array}{l}\therefore CisP.Iof\left(1\right)\&\left(2\right)\\ x y1\\ \begin{array}{cccc}-3& -15& 1& -3\\ 1& -17& 3& 1\end{array}\\ \frac{x}{51+15}=\frac{y}{-45+17}=\frac{1}{1+9}\\ \frac{x}{66}=\frac{y}{28}=\frac{1}{10}\\ \therefore x=\frac{66}{10}\Rightarrow \frac{33}{5},y=\frac{28}{10}\Rightarrow \frac{14}{5}\\ \therefore C(\frac{33}{5},\frac{14}{5})\end{array}$