I believe the question can be summarized as
$$F[x]\equiv\Pr[X<x],\,F[x_{10},x_{50},x_{90}]=[0.1,0.5,0.9],\,x_{50}<\bar{x}<x_{90}\implies{F}[\bar{x}]=?$$
As noted by many responses, there is not enough information here to solve the problem uniquely.
However building on the OP suggestion of linear interpolation, a simple approach would be to just extend the order of the polynomial interpolation. So for example we might assume
$$F[t]=a+bt+ct^2 \,,\,t=\frac{x-x_{50}}{x_{90}-x_{50}}$$
where the coefficients must satisfy the constraints $F_0=0.5$ and $F_1=0.9$. From the definition of a CDF and PDF, we must also have $F^\prime[t]\geq{0}$.
Together these constraints give
$$a=F_0\,,\,c=F_1-F_0-b\,,\,0\leq{b}\leq{2}(F_1-F_0)$$
so we have
$$t^2\leq\frac{F[t]-F_0}{F_1-F_0}\leq{t}(2-t)$$
Note that $t\in[0,1]$, and in the linear case the expression in the middle reduces to $t$, so this includes the OP's linear interpolation solution as a special case.
If the OP was intended to be on the "tail", in the sense that $c\leq{0}$ so the PDF is decreasing over the interval, then the lower bound above can be increased from $t^2$ up to $t$ (while the upper bound stays the same, at $t(2-t)$). This is how I interpret the OP comment "in this region a linear interpolation will give a lower value".
Again I will end with the caveat that this is not a general solution, as there is no reason in general that a quadratic approximation will be appropriate for an arbitrary CDF of the $[x_{50},x_{90}]$ interval!