7. Given $P_{1}\left(x_{1}, y_{1}\right), P_{2}\left(x_{2}, y_{2}\right)$ are two points on the graph of the quadratic function $y=a x^{2}+b x+c(a b c \neq 0)$ that are symmetric with respect to the axis of symmetry. Then when $x=x_{1}+x_{2}$, the value of $y$ is ( ).
(A) 0
(B) $\frac{4 a c-b^{2}}{4 a}$
(C) $c$
(D) Cannot be determined
7. C

By symmetry, we know $y_{1}=y_{2}$, that is,
$$
a x_{1}^{2}+b x_{1}+c=a x_{2}^{2}+b x_{2}+c .
$$

Rearranging gives $a\left(x_{1}+x_{2}\right)+b=0, x_{1}+x_{2}=-\frac{b}{a}$. Therefore,
$$
y=a\left(-\frac{b}{a}\right)^{2}+b\left(-\frac{b}{a}\right)+c=c .
$$