I'll use your example of $(x_1,y_1)=(10,25)$ and $(x_2,y_2)=(42,79)$ to illustrate the idea. This is merely a reformulation of the answers given so far, but this might be a bit more convenient.
To get a few things we need out of the way, we compute the midpoint $(x_c,y_c)=(26,52)$, the segment length $d=2\sqrt{985}$, and the slope $m=\dfrac{27}{16}$.
The line perpendicular to our given segment has the slope $-\dfrac1{m}=-\dfrac{16}{27}$ and passes through $(x_c,y_c)$. A particularly convenient parametrization of this line is
$\begin{align*}x&=x_c+\frac{mu}{\sqrt{1+m^2}}\\y&=y_c-\frac{u}{\sqrt{1+m^2}}\end{align*}$
(See the stuff below the dividing line for a derivation.)
To get any of the two needed points we replace $u$ in the line parametrization with $u=\pm d\dfrac{\sqrt 3}{2}=\pm\sqrt{2955}$; this yields the points $(26\pm27\sqrt 3,52\mp16\sqrt 3)$.
Deriving a convenient parametrization for the line
Start with the usual point-slope form of the line:
$y-y_1=m(x-x_1)$
It is known that if the line makes an angle $\theta$ with respect to the horizontal axis, then $m=\tan\,\theta$. We can thus also consider the alternative form
$\frac{y-y_1}{\sin\,\theta}=\frac{x-x_1}{\cos\,\theta}$
Equate the two sides of this equation to a new parameter $u$, yielding
$\begin{align*}x&=x_1+u\cos\,\theta\\y&=y_1+u\sin\,\theta\end{align*}$
To remove $\theta$, we use the relation $\theta=\arctan\,u$, and express the cosine and sine entirely in terms of the tangent:
$\begin{align*}\cos\,\theta&=\frac1{\sqrt{1+\tan^2 \theta}}\\\sin\,\theta&=\frac{\tan\,\theta}{\sqrt{1+\tan^2 \theta}}\end{align*}$
This finally yields
$\begin{align*}x&=x_1+\frac{u}{\sqrt{1+m^2}}\\y&=y_1+\frac{um}{\sqrt{1+m^2}}\end{align*}$
The "convenient" part is due to the fact that $u$ can be interpreted as (signed) distance; that is, if $u=\pm h$, then the length of the segment joining the resulting point and $(x_1,y_1)$ is $h$. The sign determines any of the two directions one can proceed from the starting point.
A similar formula for the perpendicular is conveniently realized by replacing $m$ with $-\dfrac1{m}$.