How do I prove that $x,y$ and $z$ lie on the same line if they are related to the same parameter?

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How exactly does that work? I would be grateful for a proof but barring that just the name of the theorem, so I can look it up myself.

My work/thoughts so far:

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If we start with a system like the one below:

$$y+az=b$$ $$x+fz=g$$

I guess I could solve both sides for z and get something like:

$$z=\frac{b-y}{a}$$ $$z=\frac{g-x}{f}$$

Which I guess implies that: $$\frac{b-y}{a}=\frac{g-x}{f} $$ $$y=\frac{a(g-x)}{f}+b $$

Which is really just another way of writing $y=kx+m$, so for every value of z, the relationship between x and y can be written as a line. Since there is an infinite number of z, that means we can vizualize all the possible lines as an infinite plane.

If we could define a vectpr z we could restrict the possible solutions to the values in the plane that satisfiy both $y=\frac{a(g-x)}{f}+b $ and the equation for z. It just so happens we have two expressions for z, $z=\frac{b-y}{a}$ and $z=\frac{g-x}{f}$.

I dont' know, where do I go from here?

Answer 1

I’m awfully sorry, but your question is imprecise, especially since I don’t see mention of any parameter anywhere in the title or body. So my attempt at an answer may not satisfy you at all. Nonetheless:

It seems to me that $x$, $y$, and $z$ are to be the coordinates of points in $3$-space $\Bbb R^3$. And that $a$, $b$, $f$, and $g$ are parameters which define the problem. You pretty clearly are of the same opinion.

Furthermore, you have correctly solved for $z$ in the two cases, $z=-\frac1ay+\frac ba$, $z=-\frac1fx+\frac gf\,$. These say, separately, that a choice of $x$ and $y$ gives you $z$. But not necessarily the same $z$ in the two cases. For instance, say $x=1$ and $y=2$. Then you get, in the first case, the point $(1,2,-\frac2a+\frac ba$) and in the second case $(1,2,\frac1f+\frac gf)$. As you see, the $z$-values need not be the same.

There’s where you went wrong, when you said, “Which I guess implies that…”

You want to think of the situation geometrically, where the first equation defines a particular plane in space (depending on the values of $a$ and $b$), and the second equation defines some other plane in space. (They will not be parallel planes in this case, because in the two cases, the vectors perpendicular are $(0,1,a)$ and $(1,0,f)$, definitely not parallel.) Since the planes are not parallel, they have a line in common. (If your geometric intuition is still in its infancy, think of the east wall of your room and the south wall, which intersect in a vertical line, namely the SE seam between the two walls.)

If you want a parametric representation of this line, the work is done for you. You have the three parametric formulas, in terms of a parameter $t$:

\begin{align} x&=g-ft\\ y&=b-at\\ z&=t \end{align}