6. Given point $A(1,2)$, a line passing through point $(5,-2)$ intersects the parabola $y^{2}=4 x$ at two other points $B, C$. Then, $\triangle A B C$ is
(A) an acute triangle
(B) an obtuse triangle
(C) a right triangle
(D) the answer is uncertain
(C)
6.【Analysis and Solution】Let $B\left(t^{2}, 2 t\right), C\left(s^{2}, 2 s\right), s \neq t, s \neq 1, t \neq 1$. Then the equation of line $B C$ is $\frac{y-2 t}{2 s-2 t}=$ $\frac{x-t^{2}}{s^{2}-t^{2}}$, simplifying, we get $2 x-(s+t) y+2 s t=0$.
Since line $B C$ passes through the point $(5,-2)$, we have $2 \times 5-(s+t)(-2)+2 s t=0$, which simplifies to $(s+1)(t+1)=4$.
Therefore, $k_{A B} \cdot k_{A C}=\frac{2 t-2}{t^{2}-1} \cdot \frac{2 s-2}{s^{2}-1}=\frac{4}{(s+1)(t+1)}=-1$,
Thus, $\angle B A C=90^{\circ}$, meaning $\triangle A B C$ is a right triangle.