#math/calculus
The distance between $(x_{1},y_{1})$ and $(x_{2},y_{2})$ is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$. When $x_{1}\neq x_{2}$, the slope of the line through these points is
$
m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{\Delta y}{\Delta x}.
$
The common forms of a nonvertical line are
$
\begin{aligned}
y-y_{1}&=m(x-x_{1}) &&\text{point-slope},\\
y&=mx+b &&\text{slope-intercept},\\
y-y_{1}&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1}) &&\text{two-point}.
\end{aligned}
$
```tikz
\begin{document}
\begin{tikzpicture}[scale=0.9,line cap=round]
\draw[->,black!45] (-0.3,0) -- (5.2,0) node[right] {$x$};
\draw[->,black!45] (0,-0.3) -- (0,3.7) node[above] {$y$};
\coordinate (P) at (1,0.8);
\coordinate (Q) at (4,3.0);
\draw[very thick,blue] (0.2,0.21) -- (4.7,3.51);
\draw[dashed,teal] (P) -- (4,0.8) -- (Q);
\fill (P) circle (1.8pt) node[above left] {$(x_1,y_1)$};
\fill (Q) circle (1.8pt) node[above right] {$(x_2,y_2)$};
\node[below] at (2.5,0.8) {$\Delta x$};
\node[right] at (4,1.9) {$\Delta y$};
\end{tikzpicture}
\end{document}
```
> [!Insight]
> Point-slope form says: start at $(x_{1},y_{1})$, move horizontally by $x-x_{1}$, and change height by $m(x-x_{1})$.
A vertical line has equation $x=c$ and undefined slope. A horizontal line has equation $y=c$ and slope $0$.
> [!Practice]
> Marsden & Weinstein, p. 34, exercise 44.