Clean up typos and run-on sentences. (#300)

* Clean up typos and run-on sentences.

* Make requested changes

docs/chapters/abc/content.en-GB.md

@@ -4,9 +4,9 @@ De Casteljau's algorithm is the pivotal algorithm when it comes to Bézier curve
 
 How does that work? Succinctly: we run de Casteljau's algorithm in reverse!
 
-In order to run de Casteljau's algorithm in reverse, we need a few basic things: a start and end point, a point on the curve that want to be moving around, which has an associated *t* value, and a point we've not explicitly talked about before, and as far as I know has no explicit name, but lives one iteration higher in the de Casteljau process then our on-curve point does. I like to call it "A" for reasons that will become obvious.
+In order to run de Casteljau's algorithm in reverse, we need a few basic things: a start and end point, a point on the curve that we want to be moving around, which has an associated *t* value, and a point we've not explicitly talked about before, and as far as I know has no explicit name, but lives one iteration higher in the de Casteljau process then our on-curve point does. I like to call it "A" for reasons that will become obvious.
 
-So let's use graphics instead of text to see where this "A" is, because text only gets us so far: move the sliders for the following graphics to see what, given specific `t` value, our `A` coordinate is. As well as some other coordinates, which taken together let us derive a value that the graphics call "ratio": if you move the curve's points around, A, B, and C will move, what happens to that value?
+So let's use graphics instead of text to see where this "A" is, because text only gets us so far: move the sliders for the following graphics to see what, given a specific `t` value, our `A` coordinate is. As well as some other coordinates, which taken together let us derive a value that the graphics call "ratio": if you move the curve's points around, A, B, and C will move, what happens to that value?
 
 <div class="figure">
 
@@ -29,13 +29,13 @@ So these graphics show us several things:
 
 These three values A, B, and C allow us to derive an important identity formula for quadratic and cubic Bézier curves: for any point on the curve with some `t` value, the ratio of distances from A to B and B to C is fixed: if some `t` value sets up a C that is 20% away from the start and 80% away from the end, then _it doesn't matter where the start, end, or control points are_; for that `t` value, `C` will *always* lie at 20% from the start and 80% from the end point. Go ahead, pick an on-curve point in either graphic and then move all the other points around: if you only move the control points, start and end won't move, and so neither will C, and if you move either start or end point, C will move but its relative position will not change.
 
-So, how can we compute `C`? We start with our observation that `C` always lies somewhere between the start and ends points, so logically `C` will have a function that interpolates between those two coordinates:
+So, how can we compute `C`? We start with our observation that `C` always lies somewhere between the start and end points, so logically `C` will have a function that interpolates between those two coordinates:
 
 \[
   C = u(t) \cdot P_{start} + (1-u(t)) \cdot P_{end}
 \]
 
-If we can figure out what the function `u(t)` looks like, we'll be done. Although we do need to remember that this `u(t)` will have a different for depending on whether we're working with quadratic or cubic curves. [Running through the maths](https://mathoverflow.net/questions/122257/finding-the-formula-for-bezier-curve-ratios-hull-point-point-baseline) (with thanks to Boris Zbarsky) shows us the following two formulae:
+If we can figure out what the function `u(t)` looks like, we'll be done. Although we do need to remember that this `u(t)` will have a different form depending on whether we're working with quadratic or cubic curves. [Running through the maths](https://mathoverflow.net/questions/122257/finding-the-formula-for-bezier-curve-ratios-hull-point-point-baseline) (with thanks to Boris Zbarsky) shows us the following two formulae:
 
 \[
   u(t)_{quadratic} = \frac{(1-t)^2}{t^2 + (1-t)^2}
@@ -47,7 +47,7 @@ And
   u(t)_{cubic} = \frac{(1-t)^3}{t^3 + (1-t)^3}
 \]
 
-So, if we know the start and end coordinates, and we know the *t* value, we know C, without having to calculate the `A` or even `B` coordinates. In fact, we can do the same for the ratio function: as another function of `t`, we technically don't need to know what `A` or `B` or `C` are, we can express it was a pure function of `t`, too.
+So, if we know the start and end coordinates and the *t* value, we know C without having to calculate the `A` or even `B` coordinates. In fact, we can do the same for the ratio function. As another function of `t`, we technically don't need to know what `A` or `B` or `C` are. It, too, can be expressed as a pure function of `t`.
 
 We start by observing that, given `A`, `B`, and `C`, the following always holds:
 
@@ -67,7 +67,7 @@ And
   ratio(t)_{cubic} = \left | \frac{t^3 + (1-t)^3 - 1}{t^3 + (1-t)^3} \right |
 \]
 
-Which now leaves us with some powerful tools: given thee points (start, end, and "some point on the curve"), as well as a `t` value, we can _construct_ curves: we can compute `C` using the start and end points, and our `u(t)` function, and once we have `C`, we can use our on-curve point (`B`) and the `ratio(t)` function to find `A`:
+Which now leaves us with some powerful tools: given three points (start, end, and "some point on the curve"), as well as a `t` value, we can _construct_ curves. We can compute `C` using the start and end points and our `u(t)` function, and once we have `C`, we can use our on-curve point (`B`) and the `ratio(t)` function to find `A`:
 
 \[
   A = B - \frac{C - B}{ratio(t)} = B + \frac{B - C}{ratio(t)}
@@ -82,7 +82,7 @@ With `A` found, finding `e1` and `e2` for quadratic curves is a matter of runnin
     \end{aligned} \right .
 \]
 
-And then reverse engineer the curve's control control points:
+And then reverse engineer the curve's control points:
 
 \[
     \left \{ \begin{aligned}
@@ -91,4 +91,4 @@ And then reverse engineer the curve's control control points:
     \end{aligned} \right .
 \]
 
-So: if we have a curve's start and end point, then for any `t` value we implicitly know all the ABC values, which  (combined with an educated guess on appropriate `e1` and `e2` coordinates for cubic curves) gives us the necessary information to reconstruct a curve's "de Casteljau skeleton". Which means that we can now do several things: we can "fit" curves using only three points, which means we can also "mold" curves by moving an on-curve point but leaving its start and end point, and then reconstructing the curve based on where we moved the on-curve point to. These are very useful things, and we'll look at both in the next few sections.
+So: if we have a curve's start and end points, then for any `t` value we implicitly know all the ABC values, which  (combined with an educated guess on appropriate `e1` and `e2` coordinates for cubic curves) gives us the necessary information to reconstruct a curve's "de Casteljau skeleton". Which means that we can now do several things: we can "fit" curves using only three points, which means we can also "mold" curves by moving an on-curve point but leaving its start and end points, and then reconstruct the curve based on where we moved the on-curve point to. These are very useful things, and we'll look at both in the next few sections.