6. Given point $A(1,2)$, a line passing through point $(5,-2)$ intersects the parabola $y^{2}=4 x$ at two other points $B$ and $C$. Then, $\triangle A B C$ is
A. Acute triangle
B. Obtuse triangle
C. Right triangle
D. The answer is uncertain
Given $B C: x=n(y+2)+5$, combining $\left\{\begin{array}{l}y^{2}=4 x, \\ x=n(y+2)+5\end{array}\right.$ $\Rightarrow y^{2}-4 n y-8 n-20=0$. Let $B\left(x_{1}, y_{1}\right), C\left(x_{2}, y_{2}\right)$, then $y_{1}+y_{2}=4 n, y_{1} y_{2}=-4(2 n+5)$.
Thus, $\overrightarrow{A B} \cdot \overrightarrow{A C}=\left(x_{1}-1\right)\left(x_{2}-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)$ $=\left(\frac{y_{1}^{2}}{4}-1\right)\left(\frac{y_{2}^{2}}{4}-1\right)+\left(y_{1}-2\right)\left(y_{2}-2\right)=\frac{\left(y_{1}-2\right)\left(y_{2}-2\right)}{16}\left[y_{1} y_{2}+2\left(y_{1}+y_{2}\right)+20\right]$ $=\frac{\left(y_{1}-2\right)\left(y_{2}-2\right)}{16}[-4(2 n+5)+8 n+20]=0 \Rightarrow A B \perp A C$.
Therefore, $\triangle A B C$ is a right triangle. So the answer is $C$.