7. The line $x-2 y-1=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$, and $C$ is a point on the parabola such that $\angle A C B=90^{\circ}$. Then the coordinates of point $C$ are $\qquad$.
Let $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right), C\left(\frac{t^{2}}{4}, t\right)$, and solve the system $\left\{\begin{array}{l}y^{2}=4 x, \\ x=2 y+1\end{array} \Rightarrow y^{2}-8 y-4=0\right.$. Then $y_{1}+y_{2}=8, x_{1}+x_{2}=2\left(y_{1}+y_{2}\right)+2=18, y_{1} y_{2}=-4, x_{1} x_{2}=\frac{y_{1}^{2} y_{2}^{2}}{16}=1$.
Thus, $\overrightarrow{C A} \cdot \overrightarrow{C B}=\left(\frac{y_{1}^{2}}{4}-\frac{t^{2}}{4}\right)\left(\frac{y_{2}^{2}}{4}-\frac{t^{2}}{4}\right)+\left(y_{1}-t\right)\left(y_{2}-t\right)$ $=\left(y_{1}-t\right)\left(y_{2}-t\right)\left[\frac{1}{16}\left(y_{1}+t\right)\left(y_{2}+t\right)+1\right]=0 \Rightarrow\left(y_{1}+t\right)\left(y_{2}+t\right)+16=0$ $\Rightarrow y_{1} y_{2}+t\left(y_{1}+y_{2}\right)+t^{2}+16=0 \Rightarrow t^{2}+8 t+12=0 \Rightarrow t=-2$ or -6. In conclusion, the coordinates of point $C$ are $(1,-2)$ or $(9,-6)$.