5. $x+1=-(2 y-1)^{2}$.
Since $a, b, c$ form an arithmetic sequence, then $a-2 b+c=0$. Therefore, point $P_{0}(-2,1)$ lies on the line $b x+a y+c=0$,

and $P_{0}$ is also on the parabola $y^{2}=-\frac{1}{2} x$.
Let $P_{1}\left(x_{1}, y_{1}\right)$ be the other intersection point of the line and the parabola, then the midpoint $P(x, y)$ of $P_{0} P_{1}$ satisfies
$$
x=\frac{1}{2}\left(x_{1}-2\right), \quad y=\frac{1}{2}\left(y_{1}+1\right),
$$

i.e., $x_{1}=2 x+2, y_{1}=2 y-1$.
Since $y_{1}^{2}=-\frac{1}{2} x_{1}$, thus, $x+1=-(2 y-1)^{2}$.
5. Given that $a, b, c$ are three real numbers in an arithmetic sequence. Then the locus of the midpoint of the chord formed by the intersection of the line $b x+a y+c=0$ and the parabola $y^{2}=-\frac{1}{2} x$ is $\qquad$ .