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In school, I have been learning that the standard expression of a linear equation in two variables is of the form: $ax + by + c = 0$ while $a \neq 0$ and $b \neq 0$. I want to understand the purpose of the constant 'c' in this equation and where it was derived from, whether it be graphically or algebraically.

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One can't derive a definition out of nothing; one can only justify why it is a useful definition, as the currently posted answers do. Alternatively, if you already have a different definition of "linear equation in two variables" that you would like to derive the expression $ax+by+c=0$ from, you should tell us what that is. I don't understand what "first principles" you have in mind. –  Rahul Feb 9 '12 at 16:19
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We can't have both $a$ and $b$ equal to $0$. But it is perfectly possible for one of $a$ or $b$ to be $0$. If $a=0$ and $b\ne 0$, we get $by+c=0$, that is, $y=-b/c$, a line parallel to the $x$-axis. If $b=0$, we get $x=-c/a$, a line parallel to the $y$-axis. –  André Nicolas Feb 9 '12 at 16:23
    
Thank you for your reply. What I was wondering is whether one can start by expressing the slope of any line by saying y/x = a/b for straight lines passing through the origin (0,0). Further for lines which intersect the x and y axis a constant 'c' to be included in the relationship. Please advice. –  Ram Sidharth Feb 9 '12 at 16:37
    
@Ram Siddarth: Good, you know about slope. Consider the line with slope $m$ that passes through $(p,q)$. Then a point $(x,y)\ne (p,q)$ is on the line if and only if $\frac{y-q}{x-p}=m$. For $x\ne p$ this is equivalent to $y-q=m(x-p)$, which simplifies to $mx+(-1)y= mp-q$. This is of the right shape. Conversely, one can verify that (except if $b=0$), if $(p,q)$ and $(r,s)$ are solutions of $ax+by=c$, then $\frac{s-q}{r-p}=-a/b$. This says that the points that satisfy the equation lie on a line. The $c$ by itself has no geometric meaning: we can always multiply through by a non-zero $k$. –  André Nicolas Feb 9 '12 at 17:17
    
Thanks a lot Andre. –  Ram Sidharth Feb 9 '12 at 17:27
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3 Answers

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Not quite sure what you're asking about fundamental principles. Do you mean more or less from the definition of a line? Well, if you define a line as having constant slope, you can write this as

$\frac{y-y_1}{x-x_1}=m$

which can then be manipulated to

$y-y_1=m(x-x_1),y=mx+y_1-mx_1$

If you plug in the point $(0,b)$ for $(x_1,y_1)$, you get the slope-intercept form

$y=mx+b$

From here, you could subtract both sides by $mx$ and multiply by a constant to clear fractions if you wish to get the form you desire.

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Thank you Mike. The usage of 'fundamental principles' may have been inappropriate. However, the above derivation is in line with what I was thinking. –  Ram Sidharth Feb 9 '12 at 16:54
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Let's start with the fact that we want linear equations in two variables to represent a line.

Suppose we didn't have a constant $c$. If we considered all equations of the form $ax+by=0$, we would only get lines through the origin, as follows: we could solve our equation to get $y=-\frac{a}{b}x$, which always passes through $(0,0)$.

Adding the constant $c$ allows us to consider all lines in the plane. Specifically the equation $ax+by+c=0$ passes through the points $(0, -\frac{c}{b})$ and $(-\frac{c}{a},0)$

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Thanks a lot for the explanation, but please advice me if it is possible to derive the linear equation from fundamental principles? –  Ram Sidharth Feb 9 '12 at 16:05
    
What are the fundamental principles you would like to start with? (For instance, the intention of having the linear equation describe a line is a fundamental principle in my mind) –  Aru Ray Feb 9 '12 at 18:18
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$ax+by=0$ is an equation of a line passing through the origin. This can be verified by putting $x=y=0$ in the equation. LHS=RHS and hence verified. If however, you want an equation of a line which doesn't pass through the origin, the factor 'c' is needed. In the more 'popular' form of the equation $y = mx+c$ (Just a fancy way of arranging $ax+by+c=0$ to include slope and intercept), c becomes the intercept of the line on the Y axis. In other words, when $x=0$, $y = c$. The factor c only adds to "move" the line parallel to itself up or down. See below. There are 2 lines parallel to each other (Lets call them upper and lower). One of them is $y = 2x + 3 (Upper)$ and the other $y = 2x (Lower)$ The factor 3 "moves" the line up by 3 units when the line intersects the Y axis. 2 Parallel Lines

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Thanks a lot for the explanation, but please advice me if it is possible to derive the linear equation from fundamental principles? –  Ram Sidharth Feb 9 '12 at 16:05
    
What linear equation? What first principles? Basic algebra of Real numbers states that numbers can be multiplied, added and subtracted to give another real number. ax is basically a times x which is "scaling" of a number. Similar is by. They can be added to give ax + by –  Inquest Feb 9 '12 at 16:29
    
Thank you for your reply. What I was wondering is whether one can start by expressing the slope of any line by saying y/x = a/b for straight lines passing through the origin (0,0). Further for lines which intersect the x and y axis a constant 'c' to be included in the relationship. Please advice. – –  Ram Sidharth Feb 9 '12 at 16:38
    
For a linear curve (A line), the slope is constant and is equal to the difference in the y co-ordinates of two points to the difference between the x-coordinates of the same 2 points (m = $\frac{\Delta y}{\Delta x}$) This is true for all lines. Incidentally, for lines passing through the origin, one of the points can be chosen as (0,0) hence slope is $\frac{y}{x}$. For other lines, m = Slope = $\frac{y_2-y_1}{x_2-x_1}$. Thus, $y_2 - y_1 = m(x_2-x_1)$ This can be arranged in the form you wanted. –  Inquest Feb 9 '12 at 16:52
    
Thank you Nunoxic. –  Ram Sidharth Feb 9 '12 at 16:57
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