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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

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  • $\begingroup$ If you want a formal proof you have to start from a formal definition of straight line. What is your definition of straight line? $\endgroup$ – Emilio Novati Oct 4 '16 at 19:55
  • $\begingroup$ Incidentally, 1) is false, take the equation $0x+0y=0$. $\endgroup$ – Rene Schipperus Oct 4 '16 at 20:00
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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.

enter image description here

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  • $\begingroup$ so we can only come to this conclusion through observations? There are no hard proves? $\endgroup$ – MartianCactus Oct 5 '16 at 3:28
  • $\begingroup$ I thought you were looking for an intuitive grasp. As I don't know your level of studies, and even less in which educational system you have been raised, I assumed that you had been presented a proof of the equivalence between this geometrical fact (aligned points) and the algebraic fact (ax+by=c). I am going to make a little edit to my answer in this direction. $\endgroup$ – Jean Marie Oct 5 '16 at 5:41
  • $\begingroup$ @MartianCactus Edit done. Say if you consider it as a "hard" proof... $\endgroup$ – Jean Marie Oct 5 '16 at 6:31
  • $\begingroup$ sorry its been a while..I haven't finished reading it completely as I am stuck- "It will suffice now to increase x by 2 while decreasing simultaneously y by 1 for preserving the balance (13)."- I dont know how this will preserve the balance.. $\endgroup$ – MartianCactus Oct 10 '16 at 6:59
  • $\begingroup$ The "balance" is $13$. We have the geometric progressions $x=\cdots3\rightarrow5\rightarrow7\rightarrow9\cdots$, and $y=\cdots5\rightarrow4\rightarrow3\rightarrow2\cdots$, that you combine by doing $x+2y$ and you observe that this quantity remains constant with value $13$. $\endgroup$ – Jean Marie Oct 10 '16 at 7:23

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