Co - ordinates of a point lying on the perpendicular bisector of a segment. Having two points $A(xa, ya)$ and $B(xb, yb)$ and knowing a value $k$ representing the length of a perpendicular segment in the middle of $[AB]$, how can I find the other point of the segment?


The known values are $xa$, $ya$, $xb$, $yb$. Also, it's obvious that $xm = \frac{(xa + xb)}{2}$ and $ym = \frac{(ya + yb)}{2}$

How to find $N(xn, yn)$?
 A: Let $\overrightarrow{AB} = (xb-xa,\,yb-ya)$, and $\vec p\perp \overrightarrow{AB}$. We can get $\vec p = \big(yb-ya,\, -(xb-xa)\big)$. Now, let $\vec n = \frac{\vec p}{|\vec p|}=(n_x,\, n_y)$ be a unit vector. And now you have $\overrightarrow{OM} + k\vec n = \overrightarrow{ON}$ ($O$ is the origin), or
$$
xn = xm + kn_x,\\
yn = ym + kn_y.
$$
A: The segment $NM$ defines the perpendicular bisector of $AB$. That means that $NA = NB$, which  implies $(NA)^2 = (NB)^2 \tag{1}.$ 
By $(1)$, we have: 
$$(x_n - x_a)^2 + (y_n - y_a)^2 = (x_n - x_b)^2 + (y_n - y_b)^2,$$ which yields an equation of the form $$\alpha x_n + \beta y_n = 
\gamma,\, \quad \alpha,\beta,\gamma \in \mathbb R.\tag{2}$$
Let $M(x_m, y_m)$ be the midpoint of $AB$. Then we want:
$$(x_n - x_m)^2  + (y_n - y_m)^2 = k^2\tag{3}.$$
Combining $(2)$ and $(3)$ will lead us to a quadratic equation of $x_n$ (or $y_n$, it depends on the substitution we make), whence we take $2$ values for $x_n$, e.g $x_{n_1}, x_{n_2}$. By $(2)$, we can find the respective values $y_{n_1}$ and $y_{n_2}$.
Basically, we have $2$ points: $N(x_{n_1}, y_{n_1})$ and $N'(x_{n_2}, y_{n_2})$, which are actually symmetric with respect to the line $AB$. Both  $N,N'$ satisfy all the above equations. 
