重心坐标系中直线

Equation of lines

Credits to this article.

What makes barycentric more useful is that the equation of line is very simple.

The equation of a line is $ux + vy + wz = 0$ where $u, v, w$ are reals. (which are unique up to scaling.)

So, what is the equation of a line passing through a vertex, you might ask. We can make use of the equation above.

Great, you remembered. So, substituting this into the equation of line gives $u = 0$. So, $vy + wz = 0 \Rightarrow y = \frac{-w}{v}z \Rightarrow y = kz$, for some constant $k$.

It's $z = 0$.

As an exercise, try to use the techniques you'd learned so far to prove Ceva's and Menelaus's Theorem.

Basics of lines in barycentric coordinates

So it should be understood how barycentric coordianates are related to areas now. As a recap, consider the following diagram:

Now from points, we move on to lines. How would we describe lines using barycentric coordinates? For example, consider the line $BC$.

So far, we know that similarly, the lines $BC, AC, AB$ would correspond to the equations $x=0, y=0, z=0$. But what about the general case? What would the general equation of a line be?

In the Cartesian plane, we know that a line can be represented as the equation $ux+vy+w=0$, where $u,v,w$ are constants. Why not we try to do the same? Instead of the $w$, we replace that with $wz$, i.e. $ux+vy+wz=0$.

Now you may ask, why is this so? Well the intuitive answer is to consider the nature of lines. If you have two lines, they either intersect at a single point, never intersect, or coincide (loosely speaking, intersecting at an infinite number of points). Hmm. This sounds like a set of equations having one set of solution, no solutions, or an infinite set of solutions. Rings a bell? Yes, these are the nature of the solutions to a system of linear equations, therefore the equation to describe a line would be linear (or actually 'linear' literally means 'created by lines').

So how do we find these constants? Well since two points determine a line, suppose we the two points are $P(x_1:y_1:z_1), Q(x_2:y_2:z_2)$, we just need to solve the system of $$\left\{ \begin{array}{lr} ux_1+vy_1+wz_1 & = 0 \\ ux_2+vy_2+wz_2 & = 0 \end{array} \right.$$ in $u,v,w$. A note before we go on: any triple of reals $\{u,v,w\}$ would give us a line, as long they are not all zero. Now let's try our previous example again, with points $B,C$ being our two points defining the line $BC$. Have some time thinking how to solve these two equations. Do you run into any trouble?

Immediately you would arrive at $v=0, w=0$, and have no clue about what $u$ would be. In fact, $u$ then can be anything (all nonzero reals)! $$x=0, -x=0, 2.5x=0, \sqrt[2014]{2014}x=0$$ just all mean the same thing/equation/line!

We should investigate further. Consider the midpoint of line $BC$, and call it $M$.

By the knowledge of the line $BC$ equivalent to the equation $x=0$, we know that $M$ must be in the form $(0:?:??)$. Another recap!

You know what to do now! So now we need to solve the system of equations: $$\left\{ \begin{array}{lr} u(1)+v(0)+w(0) & = 0 \\ u(0)+v(0.5)+w(0.5) & = 0 \end{array} \right.$$ in $u,v,w$.

Now let us make some observations from the previous findings. Firstly, consider the special case when a line passes through one of the vertices, say $C$.

Suppose there is a point $S$ on segment $AB$ that divides the segment into the ratio $AS:SB = m:n$.

Nice, so we now know the general equation of a cevian. What about any line? Let's revisit the general case. $$\left\{ \begin{array}{lr} ux_1+vy_1+wz_1 & = 0 \\ ux_2+vy_2+wz_2 & = 0 \end{array} \right.$$ Remember how we wanted to solve this in general for $u,v,w$? If we solve this (you do not need to this), we should get $$u=k(y_1z_2-y_2z_1), v=k(z_1x_2-z_2x_1), w=k(x_1y_2-x_2y_1)$$ where $k$ is a nonzero real. This tells us that $u,v,w$ exists uniquely up to scaling, that is, they can be all multiplied by some real. This makes very handy in a sense that we may use un-homogenized coordinates rather than homogenized coordinates (that is, no constraint of $x+y+z=1$. YAY!!!), when dealing with lines. What a bonus to make life easier!

Now we would substitute the values $u,v,w$ into our equation, but that is somewhat ugly. Those who are familiar with matrices would recognize this is related to the determinant of a $3\times 3$ matrix. In fact, the general equation of a line is therefore -drumroll- $$\left|\begin{array}{ccc}x_1&y_1&z_1\\ x_2&y_2&z_2 \\ x&y&z \end{array}\right|=0.$$ Now have a go with this formula!

Let the internal angle bisectors of angle $ABC$ and angle $BCA$ intersect lines $AC$ and $AB$ at points $E$ and $F$ respectively.

If you had an equation of the form $acy+abz-bcx=0$ or anything equivalent to that, it is perfectly fine. I left it as that form above because I find much more convenient later on (you can do so as well!), although they are fractions (don't scared of them). Also answer to the bonus are $\frac xa = \frac zc$ and $\frac xa = \frac yb$ (figure out which one is which yourself!). Bonus to bonus: if we equate the two equations, we should get the point of intersection, which is the incenter, which is just $(a:b:c)$.

And that is basically the basic of, well, lines! As a challenge, investigate parallel lines, Ceva's theorem and Menelaus theorem (if you know them, and you should know them!) using the above facts. It should be a lot of fun, and I would probably continue to write on that.

Side note: Yay!! First article done!!