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 $\qquad$ .
7. $(1,-2)$ or $(9,-6)$.

Let $A\left(x_{1}, y_{1}\right) 、 B\left(x_{2}, y_{2}\right) 、 C\left(t^{2}, 2 t\right)$.
From $\left\{\begin{array}{l}x-2 y-1=0, \\ y^{2}=4 x,\end{array}\right.$ we get $y^{2}-8 y-4=0$.
Then $\left\{\begin{array}{l}y_{1}+y_{2}=8, \\ y_{1} y_{2}=-4 \text {. }\end{array}\right.$
Also, $x_{1}=2 y_{1}+1, x_{2}=2 y_{2}+1$, so
$$
\left\{\begin{array}{l}
x_{1}+x_{2}=18, \\
x_{1} x_{2}=1 .
\end{array}\right.
$$

Since $\angle A C B=90^{\circ}$, we have $\overrightarrow{C A} \cdot \overrightarrow{C B}=0$.
Thus, $\left(t^{2}-x_{1}\right)\left(t^{2}-x_{2}\right)+\left(2 t-y_{1}\right)\left(2 t-y_{2}\right)=0$.
Substituting equations (1) and (2) into the above equation and simplifying, we get
$$
\begin{array}{l}
t^{4}-14 t^{2}-16 t-3=0 \\
\Rightarrow(t+1)(t+3)\left(t^{2}-4 t-1\right)=0 .
\end{array}
$$

Clearly, $t^{2}-4 t-1 \neq 0$. Otherwise, $t^{2}-2 \times 2 t-1=0$. Then, point $C$ lies on the line $x-2 y-1=0$, meaning point $C$ coincides with $A$ or $B$.
Therefore, $t_{1}=-1, t_{2}=-3$.
Hence, the required point $C(1,-2)$ or $C(9,-6)$.