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理想情况下，这将可以在下个秋季的学期中使用。

示例

以下示例的最终结果将类似于你在这里看到的内容 这里.

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\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}}
}