Help working through "vague" passage: eliminating a ratio I am beginning to read through Pedoe's Geometry: A Comprehensive Course and upon beginning a section on the affine plane, I can't help but get lost on where the author is going.
Any help on making the direction of the author more clear will be greatly appreciated. I have done some paraphrasing in order to make the passage easier to tackle:

We consider a plane in which the points are given by ordered pairs of real numbers $(a,b)$, and the line joining the points $P = (x_1,y_1)$ and $Q=(x_2,y_2)$ is defined as the set of points $R$ where $$R = \left\{ \left( \frac{k_2 x_1+k_1 x_2}{k_1+k_2} , \frac{k_2 y_1+k_1 y_2}{k_1+k_2}\right) \right\} $$ and $k_1,k_2 \in \Bbb{R}$ such that $k_1+k_2 \neq 0$. If we eliminate the ratio $k_1:k_2$, we find that the coordinates $(X,Y)$ of $R$ satisfy a linear equation of the form $$a_1X + a_2Y + a_3 = 0$$

The bold line in the quote is the one that has me thrown for a loop. I believe what the author is trying to show is that the set of points $R$ do indeed form a line. However, when I try to work it out myself I don't know how to apply the idea of eliminating the ratio $k_1:k_2$ nor do I know what kind of "eliminating" I am supposed to do; eliminate via subtracting off each side? Eliminate by dividing off each side (i.e multiply through by $k_2 / k_1$ ) ?
I figured out a way to show the set of points form a line with tedious algebra, but I want to know if there's a quicker, less obvious way, to do it using that ratio like the author mentions.
 A: I don't know precisely what Pedoe had in mind, but here's a way to see what's going on.
First, you get exactly the same set of points if you use only pairs $(k_1, k_2)$ where $k_1 + k_2 = 1$. Then you can omit the denominators. 
When $k_1= 1$ and $k_2 = 0$ you get the point $Q$. 
When $k_1= 0$ and $k_2 = 1$ $P$. When $0 < k_1 < 1$ you get the point on the line segment joining $P$ to $Q$ that's the fraction $K_1$ of the way from $P$ to $Q$ - think of it as a convex combination. When one of the $k$'s is negative you get the unbounded parts of the line through $P$ and $Q$.
To write the equation of the line in the form $aX + bY +  c = 0$ simply substitute $P$ and $Q$ in turn and solve, first finding one of the coefficients you can safely set to $1$.
A: Here's perhaps a way to think about it. It seems rather clunky to me (I much prefer the method in the answer of @EthanBolker), but my suspicion is that with the correct algebraic mindset this is supposed to be obvious.
If you divide by $k_1$ in the numerator and denominator of both fractions, and if you denote the ratio $\rho = k_2 / k_1$, you get
$$R = \left\{ \left( \frac{\rho \, x_1+ x_2}{1 + \rho} , \frac{\rho \, y_1+y_2}{1 + \rho}\right) \right\} 
$$
Think of this as two equations, one for each of the two coordinates $R = (X,Y)$:
$$
X = \frac{\rho \, x_1+ x_2}{1 + \rho} \qquad
Y = \frac{\rho \, y_1+y_2}{1 + \rho}
$$
You can eliminate $\rho$ by solving each of these two equations for $\rho$ and setting the results equal, and you get the single equation
$$\frac{x_2-x_1}{X-x_1} = \frac{y_2-y_1}{Y-y_1}
$$
which can then be rewritten in the form
$$\underbrace{(y_2-y_1)}_{a_1} \, X + \underbrace{(x_1-x_2)}_{a_2} \, Y + \underbrace{(-x_1y_2+y_1x_2)}_{a_3} = 0
$$
