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What is C in Standard Form derived from? [closed]

Writer Matthew Barrera
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For Example $9x + 16y = 72$. What is $72$ derived from? Why does dividing $72$ by $9x$ get us to the slope? Why does dividing $72$ by $16y$ get us to the $y$ intercept?

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1 Answer

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The "standard form" of a line is $Ax + By = C$ where $A$ and $B$ are not all zero. Usually we write lines in the slope intercept form $y = mx +b$. Here's how to go from standard form to slope-intercept form - I show all my students this.

From the standard form, divide by $B$ to allow the coefficient of $y$ to be $1$.

$$ \frac{A}{B}x + y = \frac{C}{B} $$

Now let's subtract $\frac{A}{B}x$ on both sides.

$$y = -\frac{A}{B}x + \frac{C}{B} $$

Now we do a thing called "equating coefficients." This process allows us to identify the slope and y-intercept of this line. The coefficient on $x$ is the slope, and the constant term the y-intercept. Hence we may identify $ m = -\frac{A}{B}$ and $b = \frac{C}{B}$.

Of course, you ask about $C$ in particular. This is a parameter to help define other pieces of the line - namely, intercepts. Changing $C$ changes the location of these intercepts.

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