Given,

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$. 

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$. 

In general, $x=a$ is a line lies on $x$ axis and parallel to $y$ axis and $y=b$ is a line lies on $y$ axis and parallel to $x$ axis.

The line will have a constant value of $x=7$,i.e, only $y$ will vary so it is parallel to $y-axis$.

We can write the equation in another form $4(x-3)+3(y+4)=120$
Now, we make a table

 We observe a patter that $x$ reduces by $3$ and $y$ increases by $4$. But both $x$ and $y$ cannot be negative or $0$. So, the value of $x=0$ and $y=0$ is ruled out. Hence, nine such positive integer solutions are possible.
Alternately, we can write the given equation as $4(x+3)+3(y-4)=120$. We get the same values of $x$ and $y$

Put $\displaystyle x=6$ and $\displaystyle y=0$ in given equation.
$\displaystyle 2x-3y=2\times 6-3\times 0$
=12-0
=12
Hence, the value of x and y that satisfy the equation are (6,0).

The line $x=8$ or a point $(8,0)$ lies on the $x$-axis whwich is parallel to $y$-axis and at a distance of $8$ units.

Any point on $x$ axis is of the form $(x,0)$ and any point on $y$ axis is of the form $(0,y)$. 

Given $|x|+|y|$

1. For $x>0$ and $y>0$ , we have $x+y$
2. For $x>0$ and $y<0$ , we have $x-y$
3. For $x<0$ and $y>0$ , we have $-x+y$
4. For $x<0$ and $y<0$ , we have $-x-y$
5. observe that the four equations will form a square