Derivation of slope of line formula
The formula for slope of a line as we know:
$y_2 - y_1/x_2 - x_1$ or just rise / run
What is the derivation for this formula? E.g. Why is it not rise times run for example?
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$\begingroup$The equation of the line is $$y=ax+b$$ Assume we know that $(x_1,y_1)$ and $(x_2,y_2)$ are two points on the line. Then the following system holds $$\begin{cases}y_1=ax_1+b\\y_2=ax_2+b\end{cases}$$ Now, this is a system of two equations in two unknowns ($a$ and $b$). Solving it (by subtracting the first equation from the second), yields $$a=\frac{y_2-y_1}{x_2-x_1}$$
Graphically
The slope formula is determined by $\dfrac{\text{rise}}{\text{run}}$. Our "rise" is $y_2-y_1$, and our "run" is $x_2-x_1$. Say you wanted to calculate the slope of two points: $(x_1, y_1)$, and $(x_2,y_2)$. If you draw a horizontal line that passes $(x_1, y_1)$ and a vertical line that passes $(x_2, y_2)$, you will see that it intersects at $(x_2, y_1)$. The vertical line is our rise. What is the distance between $(x_2, y_2)$ and $(x_2, y_1)$? Using the distance formula, which is: $$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Substituting our coordinates into the formula: $$D=\sqrt{(x_2-x_2)^2+(y_2-y_1)^2}$$ $$D=\sqrt{(y_2-y_1)^2}$$ $$D=y_2-y_1$$ Therefore, $y_2-y_1$ is our rise. Now for our run. What is the distance between $(x_1, y_1)$ and $(x_2, y_1)$? $$D=\sqrt{(x_2-x_1)^2+(y_1-y_1)^2}$$ $$D=\sqrt{(x_2-x_1)^2}$$ $$D=x_2-x_1$$ Our run is $x_2-x_1$. Therefore, $\dfrac{\text{rise}}{\text{run}}=\dfrac{y_2-y_1}{x_2-x_1}$. Remember that $m=\dfrac{\text{rise}}{\text{run}}$. $$\displaystyle \boxed{\therefore m=\dfrac{y_2-y_1}{x_2-x_1}}$$
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