$4 y= -3x+5$

$y= -\large\frac{3}{4}x +\frac{5}{4}$

Slope $= -\large\frac{3}{4}$

$\qquad =\tan \theta$

Angular momentum about the origin

$L= mvd$

$[p \times perpendicular \;distance]$

$\qquad= 2 \times 8 \times d$

$\qquad= 16 \;d$

$Y=0,x=\large\frac{5}{3}$

$Y=0, x= \large\frac{5}{3}$

In $OCB$,

$\sin \theta=\large\frac{OC}{OB}$

$\qquad= \large\frac{d}{OB}$

From $\Delta OAB, \tan \theta=\large\frac{5/4}{5/3}$

$\qquad= \large\frac{3}{4}$

=> $\sin \theta= \large\frac{3}{5}$

$\therefore \large\frac{3}{5}=\large\frac{d}{5/3}$

$d= 1 m$

$\therefore L=16 \;kg\;m^2/s$