Linear equations in 2 variables 1) Why do linear equations in 2 variables always represent a straight line graph when they are plotted? There must be a reason behind it right? 

Now, lest's say that we have 2 linear equations-
a1x + a2y = c2   AND a2x + b2y = c2
2) As we all know if a1/a2 ≠ b1/b2 then the lines intersect each other.
If a1/a2 = b1/b2 ≠ c1/c2 then the lines are parallel
And if a1/a2 = b1/b2 = c1/c2 then the lines are coincident
Again Why? What is the proof that this is true?
Please help. Thanks for the answer
 A: I will show, on an example, how and why an equation $ax+by=c$ corresponds in a natural way to a straight line by a "balancing process". Let us consider:
$$\tag{1}x+2y=13$$
Let us select two values $(x,y)$ that satisfy the system, for example $(x=3,y=5)$ (our first point $A$ on the figure), that is to say we start from identity: 
$$3+2 \times 5=13.$$
It will suffice now to increase $x$ by $2$ while decreasing simultaneously  $y$ by $1$ for preserving the balance (13). Thus, we obtain :
$$ 5+ 2 \times 4=13,$$
and then doing again the same operation:
$$7 + 2 \times 3 =13,$$
$$ \ \ \ \ 9 + 2 \times 2 =13 \ \ \text{etc.}$$
This staircase moves from point $(3,5) \rightarrow (5,4) \rightarrow (7,3) \rightarrow (9,2) \rightarrow \cdots$, are in fact moves along a line, while preserving value 13. 
Of course we have worked with integer coordinates, but nothing prevents us for example to add $2/5$ to the present value of $x$ while substracting 1/5 to the present value of $y$ ...
Edit: following a little exchange with the OP, here is a (linked) mathematical proof. I still do it on the previous example, but it has a general value.
Let us summarize what we have done: 


*

*we have started from a certain point $(x,y)=(x_0,y_0)$.

*then we have added 2 a certain number of times to the abscissa $x.$

*and substracted 1 the same number of times to the ordinate $y.$
If we call "$t$" this same "number of times", we have, for the most general point $(x,y)$ on the straight line:
$$\tag{2}\cases{x=x_0+2t\\y=y_0-1t}$$
which is called a parametric representation of the line. $t$ is sometimes called the time parameter.
Now, let us eliminate $t$ between the two equations in (2):
$$t=\frac{x-x_0}{2}=\frac{y-y_0}{-1}$$
$$\Leftrightarrow \ \ \ \ \ -(x-x_0)=2(y-y_0)$$
$$\tag{3}\Leftrightarrow \ \ \ \ \ x+2y=x_0+2y_0.$$
We recognize in (3) an expression of the desired form $ax+by=c$.
We can also observe that (3) expresses the preservation of quantity $x+2y$.
Remark: there are other proofs for the fact that a straight line has an equation of the form $ax+by=c$. One of the most interesting ones is by using determinants. Let us understand it at the light of (3). In fact (3) can be written under the form:
$$\cases{x-x_0=2t\\y-y_0=-1t}$$
otherwise said, by a proportionality of $\vec{M_0}M\pmatrix{x-x_0\\y-y_0}$ and directional vector $\vec{V}\pmatrix{2\\-1}.$
But this condition is equivalent to the fact that the following determinant is $0$:
$$\tag{4}\begin{vmatrix}(x-x_0) & \ 2 \\ (y-y_0) &  -1 \end{vmatrix}=0$$
Expanding this expression will give, of course the same equation as before.

