Let $w$, $s$ and $y$ be vectors in $C \in \mathbb{R}^n$ and $t \in [0,1]$. The convex combination of $w$ and $s$ are as follows
$$ y = (1-t)w + ts \tag{1} $$
Consider the following optimization problem: $$ y_* = \arg\min_{y \in C} -t\langle y,s -w \rangle + \frac{1}{2} \|y - w\|_2^2 \tag{2} $$
which can be solved to get $y_*$ because the objective, i.e., $f(y) = -t\langle y,s -w\rangle + \frac{1}{2} \|y - w\|_2^2$ is convex, so the necessary and sufficient condition in order $y_*$ be the minimizer of $(2)$ is the following:
$$ \langle -t (s -w) + (y-w), y- y_* \rangle \geq 0 \,\,\,\, \forall y \in C $$
$$ \langle y - ((1-t)w +t s) , y - y_* \rangle \geq 0 \,\,\,\, \forall y \in C \tag{3}$$
Evidently, for $y_* = (1-t)w +t s$ $(3)$ holds true not only for all $y \in C$ but also for every $y$ in $\mathbb{R}^n$. However, the cost function is strongly convex so it is strictly convex, therefore, it has one global minimizer.
Question: Using $(3)$ how can we prove that, $y_*$ is indeed $(1-t)w +t s$?