Given your initial two points $P_1$ and $P_2,$ carefully write their x-y coordinates, and find the midpoint of the line segment between them, call that point $C.$ The $x$ coordinate of $C$ is half the sum of the $x$-coordinates of $P_1$ and $P_2,$ the $y$ coordinate of $C$ is half the sum of the $y$-coordinates of $P_1$ and $P_2.$ The circle centered at $C$ with radius half the distance between $P_1$ and $P_2$ passes through both points.
Any line passing through $P_1$ does one of three things:
(A) it is perpendicular to the segment $P_1 P_2,$ your task is impossible
(B) it passes directly through $P_2,$ after passing through $C$ halfway along, you are done,
(C) it meets the circle at some second point $Q,$ in which case the line segment from $P_1$ to $Q$ and the line segment from $Q$ to $P_2$ are perpendicular.
So you need to refresh your skills in finding the equation describing a circle, given center and radius. Then you need to figure out how to write the equation of a line, given the "angle" you are talking about when the line leaves $P_1.$ Finally, you need to figure out how to find the intersection point $Q.$
Notes. If I am correct about the meaning of your "angle," the following are useful. Suppose the point $P_1$ has coordinates $(x_1,y_1),$ and your directed line leaves that point at angle $\beta.$
Then, the parametrized version of the path, parameter $t,$ is
$$ x = x_1 + t \cos \beta, \; \; y = y_1 + t \sin \beta. $$
The equation version of the path is
$$ (x-x_1) \sin \beta = (y-y_1) \cos \beta. $$