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I currently started with some basic geometry and I'm already stuck at some very very basic intuition regarding finding the line between two points in a plane.

I understand $y = mx + c$ and I am able to calculate all variables. The way I calculated $c$ thus far has been by finding the slope, and use one point in the plane to find the remainder as $c$ through $y = mx + c$.

Now the textbook used the following points: $A: (-1, -1)$ and $B: (1, 2)$ which results in $y = \frac{3}{2}x+\frac{1}{2}$ and I was able to do this myself by hand.

However, a different method without using one point and a calculcated slope involves using the following equation:

$$ c = \frac{x_2y_1 - x_1y_2}{x_2-x_1} $$

But I cannot wrap my head around or find the intuition as to why I am multiplying $x_2$ with $y_1$ and subtracting $x_1$ multiplied by $y_2$.

Considering we're dividing by $x_2 - x_1$ it must have something to do with the differences in $y$. I've calculcated both products but I don't see some sort of relation.

As a test case I used a formula I just came up with: $y = 3x + 4$ and took points $C: (-2, -2)$ and $D: (4,16)$ just to have another example but I am still stuck with why I am doing this and what the products: $x_2*y_1=4*-2=-8$ and $x_1*y_2=-2*16=-32$ tell me.

$\frac{24}{6}$ obviously is $4$ which would be the correct $c$. Yet I am missing intuition and I really want to understand this. Can someone help me?

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This follows from two facts, then some algebra.

  • Fact 1: $y=mx+c$. We'll rewrite this as $c=y-mx$
  • Fact 2: $m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ for any points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ on the line

Now, any point on the line satisfies the equation in fact $(1)$; in particular, $(x_{1},y_{1})$ satisfies it. Therefore $$ \begin{align} c &= y_{1}-mx_{1} \\ &=y_{1}-\frac{y_{2}-y_{1}}{x_{2}-x_{1}}x_{1} \\ &= \frac{y_{1}(x_{2}-x_{1})-(y_{2}-y_{1})x_{1}}{x_{2}-x_{1}} \\ &=\frac{y_{1}x_{2}-y_{2}x_{1}}{x_{2}-x_{1}} \end{align} $$

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  • $\begingroup$ Thank you! This was a really concise and straight-forward answer. $\endgroup$
    – Ventus
    Feb 8 '19 at 18:54
  • $\begingroup$ You're welcome! $\endgroup$
    – pwerth
    Feb 8 '19 at 19:58
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We can rearrange the equation as follows: \begin{align*} y &= mx+c \\ c &= y - mx \\ c &= y - \left(\frac{y_2 - y_1}{x_2 - x1}\right)x \\ c &= \frac{y(x_2 - x_1)}{x_2 - x_1} - \left(\frac{y_2 - y_1}{x_2 - x1}\right)x \\ c &= \frac{yx_2 - yx_1 - xy_2 + xy_1}{x_2 - x_1} \end{align*} Now, since this equation holds true for any pair $(x,y)$ on the line, in particular it holds for the point $(x_2, y_2)$, so we get \begin{align*} c &= \frac{yx_2 - yx_1 - xy_2 + xy_1}{x_2 - x_1} \\ c &= \frac{y_2 x_2 - y_2 x_1 - x_2 y_2 + x_2 y_1}{x_2 - x_1} \\ c &= \frac{x_2 y_1 - y_2 x_1}{x_2 - x_1} \end{align*}

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