Show that halfspace is not affine.

Let us define half-space as $$C = \{x\mid a^Tx\leq b\}$$ Intuitively (or geometrically), I understand why halfspace is not affine. But while I prove that half-space is convex, it seems to hold for affine case.

Let us choose any $x_1,x_2\in C$ and $x=\theta x_1 + (1-\theta)x_2$, then $$a^Tx = a^T(\theta x_1+(1-\theta)x_2) \leq \theta b + (1-\theta)b = b$$

As far as I know, if this inequality holds for $\theta\in\mathbb R$, $C$ is affine and if for $\theta\in[0,1]$, $C$ is convex. But in the proof, both cases seem hold. Where am I wrong?

The inequality $a^Tx=a^T(\theta x_1+(1-\theta)x_2)\leq \theta b+(1-\theta)b=b$ holds if and only if when $0\leq \theta \leq 1$. If suppose, $\theta <0$ the value of $(1-\theta)$ becomes negative. We cannot really draw any conclusion about the inequality between L.H.S and R.H.S, i.e., $a^T(\theta x_1+(1-\theta)x_2)$ and $\theta b+(1-\theta)b$.
Take a pair of points $x_1$ and $x_2$ that respect $\leq b$ and they form a segment parallel to the hyperplane's normal $a$. The $\theta$ moves the resulting point $x$ formed from $x_1$ and $x_2$ on this segment to the right(if $\theta < 0$) or to the left(if $\theta > 1$). I hope it is clear that a point on this segment for some $\theta$ will not respect $\leq b$ anymore.