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 - x_1}{x_2-x_1} = \frac{x - x_1}{\Delta_x}\\
        t &= \frac{y - y_1}{y_2-y_1} = \frac{y - y_1}{\Delta_y}\\
        t &= \frac{z - z_1}{z_2-z_1} = \frac{z - z_1}{\Delta_z}
    }
    Therefore
    \equation {
        \frac{x - x_1}{Delta_x}
            &= \frac{y - y_1}{\Delta_y}
            = \frac{z - z_1}{\Delta_z}\\
                \frac{x - x_1}{x_2-x_1}
            &= \frac{y - y_1}{y_2-y_1}
            =  \frac{z - z_1}{z_2-z_1}
    }
    \hr
    \h4{Rationale}
    We rewrite \{r = r_0 + a = r_0 + t v} in terms of \{t}.
    That is
    \equation{
        x &= x_1 + t(x_2-x_1) = x_1 + t\;Delta_x\\
        t\;Delta_x  &= x - x_1 = t(x_2-x_1)\\
        t &= \frac{x - x_1}{x_2-x_1} = \frac{x - x_1}{Delta_x} \\\\
        y &= y_1 + t(y_2-y_1) = y_1 + t\;\Delta_y\\
        t\;\Delta_y  &= y - y_1 = t(y_2-y_1)\\
        t &= \frac{y - y_1}{y_2-y_1} = \frac{y - y_1}{\Delta_y} \\\\
        z &= z_1 + t(z_2-z_1) = z_1 + t\;\Delta_z\\
        t\;\Delta_z &= z - z_1 = t(z_2-z_1) \\
        t &= \frac{z - z_1}{z_2-z_1} = \frac{z - z_1}{\Delta_z}
    }
}
\!where {
    {\Delta_x} => {\colorA{\Delta_x}}
    {\Delta_y} => {\colorA{\Delta_y}}
    {\Delta_z} => {\colorA{\Delta_z}}
    {x_1} => {\colorB{x_1}}
    {y_1} => {\colorB{y_1}}
    {z_1} => {\colorB{z_1}}
}