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docs/chapters/molding/content.en-GB.md
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@@ -5,13 +5,13 @@ Armed with knowledge of the "ABC" relation, point-on-curve projection, and guest
For quadratic curve, this is a really simple trick: we project our cursor onto the curve, which gives us a `t` value and initial `B` coordinate. We don't even need the latter: with our `t` value and "wherever the cursor is" as target `B`, we can compute the associated `C`:
\[
C = u(t)_{q} \cdot Start + \left ( 1-u(t)_{q} \right ) \cdot End
C = u(t)_{q} \cdot \textit{Start} + \left ( 1-u(t)_{q} \right ) \cdot \textit{End}
\]
And then the associated `A`:
\[
A = B - \frac{C - B}{ratio(t)_{q}} = B + \frac{B - C}{ratio(t)_{q}}
A = B - \frac{C - B}{\textit{ratio}(t)_{q}} = B + \frac{B - C}{\textit{ratio}(t)_{q}}
\]
And we're done, because that's our new quadratic control point!