Answer
They will move on the $x$ line
Work Step by Step
Components of total linear momentum of the system can be defined as :
$\overrightarrow {P}_{i}=\sum \overrightarrow {p}_{i}=\overrightarrow {P_{1}}_{i}+\overrightarrow {P_{2}}_{i}=m_{1}\overrightarrow {v_{1}}_{i}+m_{2}\overrightarrow {v_{2}}_{i}=mv\cos \theta +mv\cos \theta =2mv\cos \theta (1) $ (x direction )
And
$\overrightarrow {P}_{j}=\sum \overrightarrow {p}_{j}=\overrightarrow {P_{1}}_{j}+\overrightarrow {P_{2}}_{j}=m_{1}\overrightarrow {v_{1}}_{j}+m_{2}\overrightarrow {v_{2}}_{j}=mv\sin \theta -mv\sin \theta =0$
So if their collision is completely inelastic (and the two particles stick together), then they will only move along $x$ line:
$\overrightarrow {P}_{f}=\overrightarrow {P_{i}}$
