A point $(x,y)$ is on the line between $(x_1,y_1)$ and $(x_2,y_2)$ if and only if, for some $t\in\mathbb{R}$,
$$(x,y)=t(x_1,y_1)+(1-t)(x_2,y_2)=(tx_1+(1-t)x_2,ty_1+(1-t)y_2)$$
You need to solve
$$\begin{align*}d&=\|(x_2,y_2)-(tx_1+(1-t)x_2,ty_1+(1-t)y_2)\|=\sqrt{(tx_2-tx_1)^2+(ty_2-ty_1)^2}\\
&=\sqrt{t^2}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\hspace{5pt}\Rightarrow\hspace{5pt} |t|=\frac{d}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\end{align*}$$
You will have two values of $t$. For $t>0$ this point will be in the direction to $(x_1,y_1)$ and for $t<0$ it will be in the direction away from $(x_1,y_1)$.
Geometrically speaking, all points that are equidistance from points, p and q, all lie on the perdendicular bisector of pq.
So if $p = (x1, y1)$ and $q =(x2, y2)$ are the points, the line pq has slope, $\frac {y2 - y1}{x2 - x1}$. So the slope of a perpendicular line will have slope m = $- \frac {x2 - x1}{y2 - y1}$ As the line is a bisector, it will contain the midpoint $mid = (\frac{x1+x2}{2},\frac{y1+y2}{2})$. So the perpendicular bisector is the line with the equation $y - \frac{y1+y2}{2} = m(x - \frac{x1+x2}{2}) = y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
You want to find the points (x,y) where distance((x1,y1),(x,y)) = distance((x2,y2)(x,y)) = distance((x1,y1),(x2, y2)). Where (x,y) are on the line $y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$.
So you have three equations:
$y - \frac{y1+y2}{2} = - \frac {x2 - x1}{y2 - y1}(x - \frac{x1+x2}{2})$
$\sqrt{(x1 - x)^2 + (y1 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
and $\sqrt{(x2 - x)^2 + (y2 - y)^2} = \sqrt{(x1 - x2)^2 + (y1 - y2)^2}$
There will be two possible points that satisfy the the equations.
Okay, that's hard. BUT if you are given one of the two points the other will be symmetrically placed on the other side of the midpoint.
Ex: A=(7,3), B=(6,0), and C=(14,−1).
So the midpoint of A and B is M = (6.5, 1.5). C is 14 - 6.5 = 7.5 away from M in the x direction. C is -1 - 1.5 = -2.5 away in the y directions. So the 4th point X will also be 7.5 away in the x direction and -2.5 away in the y direction. So X = (6.5 - 7.5, 1.5 - (-2.5)) = (-1, 4).
Best Answer
The square of distance between $p1$ and $p3$ is:
$$(dx^2 + dy^2)*k^2 = 300^2$$
Now you can find $k$ and then $x3$ and $y3$