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How can one get the formula of a line going through the points $(0,0)$ and $(1,7)$

Also, how to get it in the form $ax+by=c$, I am really used to $y=ax+b$

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First we need to find the slope; we get that by using the formula:

$$\displaystyle\large m = \frac{y_2-y_1}{x_2-x_1}$$

Where $m$ is the slope, and $y_n$ and $x_n$ are the corresponding coordinates to $(x_1, y_1), (x_2, y_2)$.

$$\begin{align*}m = \frac{7 - 0}{1-0}\end{align*}=\frac{7}{1}=7$$

Now we input $m$ input the formula $y-y_1=m(x-x_1)$ in order to get the equation of the line [Note: $(x_1, y_1)$ is $(0, 0)]$:

$$y - 0 = 7(x - 0)$$

Which simplifies to:

$$y = 7x$$

We turn into $ax+by=c$ like this:

$$y - 7x = 0$$ or $$-7x + y = 0$$

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Here is an answer which can be generalised to polynomials through a number of points. Suppose we want a curve (line) which passes through $(x_1, y_1)$ and $(x_2, y_2)$ and we have two functions $f_1$ and $f_2$ with $f_1(x_1)=1$, $f_1(x_2)=0$, $f_2(x_1)=0 $, $f_2(x_2)=1$ and we set $$y=y_1f_1(x) + y_2f_2(x)$$ then we have an equation for a curve which passes through both points. Now we set $$f_1(x) = \frac {x-x_2}{x_1-x_2}, \text { and }f_2(x)= \frac {x-x_1}{x_2-x_1}$$ It is then clear that the equation for $y$ is linear in $x$. A similar idea can be used for three points to give a quadratic (etc). You need functions which vanish at all but one of the points, and add linear combinations of these to get a funtion which passes through each one.

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Def: The Slope of a line is defined as follows: If $(x_1,y_1)$ and $(x_2,y_2)$ are two points then, SLOPE = $m = \frac{y_2 - y_1}{x_2 -x_1}$.

Now, the equation of a line through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by: $y_2 - y_1 = m(x_2 - x_1)$

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As all points $(x,y)$ in your line $l$ have to fullfil $ax+by=c$, from $(0,0) \in l$ we can conclude $c = 0$. Now insert $(x=1,y=2)$ in your equation and you get $x+7y = 0$ or otherwise $y = -\frac 1 7 x$.

Getting from form 1 to 2, or the other way around is just subtracting/adding ax, and dividing by b.

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The correct equation should be $y = 7x$ (note that the point $(1, 7)$ doesn't lie on the line $y = -\frac{1}{7}x$). – Michael Albanese May 30 '15 at 19:10

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