Work in progress...

Not yet ready for public consumption; please check back in a couple of months.

Ideally this will be usable for school this coming fall.

Example

The end result of the following example will be similar to what you see over here.

``` ...

\note[boxed] { \h3{Symmetric Equation of a Line} Given \equation { t &= \frac{x - x1}{x2-x1} = \frac{x - x1}{\Deltax}\ t &= \frac{y - y1}{y2-y1} = \frac{y - y1}{\Deltay}\ t &= \frac{z - z1}{z2-z1} = \frac{z - z1}{\Deltaz} } Therefore \equation { \frac{x - x1}{Deltax} &= \frac{y - y1}{\Deltay} = \frac{z - z1}{\Deltaz}\ \frac{x - x1}{x2-x1} &= \frac{y - y1}{y2-y1} = \frac{z - z1}{z2-z1} } \hr \h4{Rationale} We rewrite {r = r0 + a = r0 + t v} in terms of {t}. That is \equation{ x &= x1 + t(x2-x1) = x1 + t\;Deltax\ t\;Deltax &= x - x1 = t(x2-x1)\ t &= \frac{x - x1}{x2-x1} = \frac{x - x1}{Deltax} \\ y &= y1 + t(y2-y1) = y1 + t\;\Deltay\ t\;\Deltay &= y - y1 = t(y2-y1)\ t &= \frac{y - y1}{y2-y1} = \frac{y - y1}{\Deltay} \\ z &= z1 + t(z2-z1) = z1 + t\;\Deltaz\ t\;\Deltaz &= z - z1 = t(z2-z1) \ t &= \frac{z - z1}{z2-z1} = \frac{z - z1}{\Deltaz} } } !where { {\Deltax} => {\colorA{\Deltax}} {\Deltay} => {\colorA{\Deltay}} {\Deltaz} => {\colorA{\Deltaz}} {x1} => {\colorB{x1}} {y1} => {\colorB{y1}} {z1} => {\colorB{z1}} } ```