(7) The line $x-2 y-1=0$ intersects the parabola $y^{2}=4 x$ at points $A$ and $B$. Point $C$ is on the parabola, and $\angle A C B=90^{\circ}$. Then the coordinates of point $C$ are $\qquad$.
(7) $(1,-2)$ or $(9,-6)$ Hint: Let $A\left(x_{1}, y_{1}\right)$, $B\left(x_{2}, y_{2}\right)$, and $C\left(t^{2}, 2 t\right)$. 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 $y_{1}+y_{2}=8$, $y_{1} y_{2}=-4$.
Also, $x_{1}=2 y_{1}+1, x_{2}=2 y_{2}+1$, so
$$
\begin{array}{l}
x_{1}+x_{2}=2\left(y_{1}+y_{2}\right)+2=18, \\
x_{1} x_{2}=4 y_{1} y_{2}+2\left(y_{1}+y_{2}\right)+1=1 .
\end{array}
$$

Since $\angle A C B=90^{\circ}$, we have $\overrightarrow{C A} \cdot \overrightarrow{C B}=0$, which means
$$
\left(t^{2}-x_{1}\right)\left(t^{2}-x_{2}\right)+\left(2 t-y_{1}\right)\left(2 t-y_{2}\right)=0,
$$

which simplifies to
$$
t^{4}-\left(x_{1}+x_{2}\right) t^{2}+x_{1} x_{2}+4 t^{2}-2\left(y_{1}+y_{2}\right) t+y_{1} y_{2}=0,
$$

which further simplifies to
$$
t^{4}-14 t^{2}-16 t-3=0,
$$

or
$$
\left(t^{2}+4 t+3\right)\left(t^{2}-4 t-1\right)=0 .
$$

Clearly, $t^{2}-4 t-1 \neq 0$, otherwise $t^{2}-2 \cdot 2 t-1=0$, then point $C$ would lie on the line $x-2 y-1=0$, making point $C$ coincide with point $A$ or point $B$. Therefore, $t^{2}+4 t+3=0$, solving for $t$ gives $t_{1}=-1, t_{2}=-3$.
Thus, the coordinates of point $C$ are $(1,-2)$ or $(9,-6)$.