= Solution
Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the <Fermat principle> minimizes the length through a reflection point $P=(s,0)$:
$$
L(s)=\sqrt{(s-a)^2+h_A^2}+\sqrt{(s-b)^2+h_B^2}.
$$
Its stationary equation is
$$
L'(s)=\frac{s-a}{\sqrt{(s-a)^2+h_A^2}}+\frac{s-b}{\sqrt{(s-b)^2+h_B^2}}=0.
$$
The two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin\theta_i=\sin\theta_r$, where both angles are measured from the <normal vector> to the mirror and lie in $[0,\pi/2)$. Hence the <law of reflection> is
$$
\boxed{\theta_i=\theta_r.}
$$
For uniqueness, differentiate once more:
$$
L''(s)=\frac{h_A^2}{[(s-a)^2+h_A^2]^{3/2}}+\frac{h_B^2}{[(s-b)^2+h_B^2]^{3/2}}>0.
$$
Thus $L$ is a <strictly convex function>, its <derivative> increases from $-2$ to $2$, and it has exactly one stationary point, its <global minimum>. Explicitly,
$$
\boxed{s_* =\frac{h_Ba+h_Ab}{h_A+h_B}.}
$$
Substitution gives $(s_*-a)/h_A=(b-s_*)/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.
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