Suppose we have two points $p_1$ and $p_2$ with coordinates $(x_1,y_1)$ and $(x_2,y_2)$ on a plane with Cartesian coordinate system. The first point moves at speed $v_1$ and at angle $\alpha_1$ to $x$-axis. The second point moves at speed $v_2$ and at angle $\alpha_2$ to $x$-axis. After some time $t$ they appear in the same position, i.e. their coordinates coincide. The problem is to find $t$ and $\alpha_2$ given all other parameters.
I composed a system of equations that meets the statement: $$ \begin{cases} t\cdot v_1\cdot \cos{\alpha_1}+x_1=t\cdot v_2\cdot\cos{\alpha_2}+x_2 \\ t\cdot v_1\cdot \sin{\alpha_1}+y_1=t\cdot v_2\cdot\sin{\alpha_2}+y_2 \end{cases} $$ It's convenient to move everything to the left side: $$ \begin{cases} t\cdot (v_1\cdot \cos{\alpha_1}-v_2\cdot\cos{\alpha_2})+x_1-x_2 =0 \\ t\cdot (v_1\cdot \sin{\alpha_1}-v_2\cdot\sin{\alpha_2})+y_1-y_2 =0 \end{cases} $$ Adding some notation to make it shorter: $$ \begin{cases} t\cdot (v_1\cdot \cos{\alpha_1}-w_x)+h_x =0 \\ t\cdot (v_1\cdot \sin{\alpha_1}-w_y)+h_y =0 \end{cases} $$
How can I effectively solve this system with respect to $t$ and $\alpha_1$? Number of solutions can be from 0 to 2.
You might like to consider formulating this problem in a different way.
Let $A(x_1,y_1)$ and $B(x_2,y_2)$, in which case you can calculate $d=|AB|$. You can also calculate the angle that $AB$ makes with the horizontal.
Now let the angle between $v_1$ and $\overrightarrow{AB}$ be $\theta$, and the angle between $v_2$ and $\overrightarrow{BA}$ be $\phi$. Without loss of generality we can assume that these angles cannot both be obtuse.
The condition for collision is that the velocity components perpendicular to $AB$ must be equal, in which case $$v_1\sin \theta=v_2\sin \phi$$ from which $\alpha_1$ can be calculated.
The time to collision is then found by dividing the initial relative displacement by the relative speed in the direction of $AB$, so that $$t=\frac{d}{v_1\cos\theta+v_2\cos\phi}$$