Let at a certain instant two particles be at points $P$ and $Q$, as shown in the following figure.

Angular momentum of the system about point $P$ :

$\vec{L}_{p}=m v \times 0+m v \times d$

$=m v d.. .(i)$

Angular momentum of the system about point $Q:$ $\vec{L}_{\Omega}=m v \times d+m v \times 0$

$=m v d…(i i)$

Consider a point $R,$ which is at a distance $y$ from point $Q,$ i.e.

$QR =y$

$PR =d-y$

Angular momentum of the system about point $R:$ $\vec{L}_{R}=m v \times(d-y)+m v \times y$

$=m v d-m v y+m v y$

$=m v d\ldots(i i i)$

Comparing equations $(i),(i i),$ and $(iii)$, we get:

$\vec{L}_{P}=\vec{L}_{0}=\vec{L}_{R}\ldots(i v)$

We infer from equation ($i v$) that the angular momentum of a system does not depend on the point about which it is taken.