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I know,

$Ax + By = C$

is the equation of straight line but a different resource says that:

$y = mx + b$

is also an equation of straight line? Are they both same?

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marked as duplicate by Zev Chonoles Apr 7 '13 at 4:55

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

    
The first is called an implicit equation, and the second a function. The values $A$, $B$, $C$ are sometimes called the coordinates for the line, and the values $m$, $b$ the parameters for the line. –  ja72 Jan 23 '12 at 17:48
    
There is an other line representation with $(x,y)=(x_1,y_1)+t\,(n_x,n_y)$ derived from a point $(x_1,y_1)$ and a direction $(n_x,n_y)$. –  ja72 Jan 23 '12 at 17:50
    
$Ax+By=C$ is sometimes called general form, general linear form, or standard form; $y=mx+b$ is often called slope-intercept form. –  Isaac Jan 29 '12 at 8:16
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2 Answers

up vote 2 down vote accepted

Yes. That is, they both give the equation of a straight line and the equation of any non-vertical line can be written in either form.

If $B\ne 0$. Then you can write $Ax+By=C$ as $$ By=-Ax+C $$ and, since $B\ne0$, the above can be written $$ y=-\textstyle{A\over B}x +{C\over B}. $$

If $B=0$, the equation is $Ax=C$, which is a vertical line when $A\ne0$. In this case you can't write it in the form $y=mx+b$ (which defines a function).

On the other hand, given $y=mx+b$, you can rewrite it as $-mx+y=b$.




Note that for the equation $Ax+By=C$ with $A$ and $B$ both non-zero:

The $y$-intercept of its graph is $C/B$ and is found by taking $x=0$.

The $x$-intercept is of its graph is $C/A$ and is found by taking $y=0$.

The slope of the line is then $ {C/B-0\over 0-C/A } = -A/B$.

enter image description here

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$Ax + By = C$

$By = -Ax + C$

$y = -(A/B)x + C/B$

Let $m = -\frac{A}{B}$. Let $b = \frac{C}{B}$.

$y = mx + b$

So they are equivalent.

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This was a good idea. It helps many people to illustrate equivalency in the most blunt way possible: Proving it. –  000 Jan 26 '12 at 20:43
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