Determine,graphically,the vertices of the triangle formed by the lines $y=x$,$3y=x$,and $x+y=8$.

(O(0,0), Q(4,4), D(6,2)) The given linear equations are:
$y=x \quad (i)$
$3y=x \quad (ii)$
$x+y=8 \quad (iii)$
To find the vertices,we determine the intersection points of these lines taken in pairs.
$1$. Intersection of $(i)$ and $(ii)$:
Substituting $y=x$ into $3y=x$,we get $3x=x$,which implies $2x=0$,so $x=0$. Thus,$y=0$. The intersection point is $O(0,0)$.
$2$. Intersection of $(i)$ and $(iii)$:
Substituting $y=x$ into $x+y=8$,we get $x+x=8$,so $2x=8$,which means $x=4$. Thus,$y=4$. The intersection point is $Q(4,4)$.
$3$. Intersection of $(ii)$ and $(iii)$:
From $(ii)$,$x=3y$. Substituting this into $(iii)$,we get $3y+y=8$,so $4y=8$,which means $y=2$. Then $x=3(2)=6$. The intersection point is $D(6,2)$.
Thus,the vertices of the triangle formed by the lines are $O(0,0)$,$Q(4,4)$,and $D(6,2)$.