Line determined by two points is $\frac{x - x_0}{x_1 - x_0} = \frac{y - y_0}{y_1 - y_0} = \frac{z - z_0}{z_1 - z_0}$

Question

I am reading a textbook that says that analytical geometry tells us that the equations of the line determined by two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ can be written in the form

$$\dfrac{x - x_0}{x_1 - x_0} = \dfrac{y - y_0}{y_1 - y_0} = \dfrac{z - z_0}{z_1 - z_0}$$

Can someone please help me understand how this is true by showing how it is derived? Thank you.

Answer 1

All the points of a line are of the form $tv + p,\ t\in\mathbb R,$ where $v$ is direction vector and $p$ is any point on the line..

Let $p = (x_0,y_0,z_0)$ and $(x_1,y_1,z_1)$ be some point on the line different than $p$. That means there exists some $t_1$ such that $(x_1,y_1,z_1) = t_1v + (x_0,y_0,z_0)$, i.e. $(x_1-x_0,y_1-y_0,z_1-z_0)$ is proportional to $v$, so we can write our line as all the points of the form $$t(x_1-x_0,y_1-y_0,z_1-z_0)+ (x_0,y_0,z_0),\ t\in\mathbb R.$$

Thus, if $(x,y,z)$ is on the line, then there exists $t\in\mathbb R$ such that $$(x,y,z) = t(x_1-x_0,y_1-y_0,z_1-z_0)+ (x_0,y_0,z_0),$$

i.e.

\begin{align} x - x_0 &= t(x_1 - x_0)\\ y - y_0 &= t(y_1 - y_0)\\ z - z_0 &= t(z_1 - z_0)\tag{1} \end{align}

and therefore $$\frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0} = \frac{z-z_0}{z_1-z_0}.\tag{2}$$

Conversely, if $(2)$ holds for $(x,y,z)$, then define $t = \frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0} = \frac{z-z_0}{z_1-z_0}$ and $(1)$ follows.

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Note that the $2$-dimensional analog is $\frac{x-x_0}{x_1-x_0} = \frac{y-y_0}{y_1-y_0}$ or $y - y_0 = \frac{y_1-y_0}{x_1-x_0}(x-x_0)$.

Answer 2

It is implicitly assumed that $x_0\neq x_1$, $y_0\neq y_1$ and $z_0\neq z_1$.

The equations of the line $\mathcal{D}$ can be found by considering a point $(x,y,z)$ and writing the equivalence:

$(x,y,z)\in \mathcal{D} \Leftrightarrow \left|\matrix{x-x_0\\y-y_0\\z-z_0}\right.\text{ and }\left|\matrix{x_1-x_0\\y_1-y_0\\z_1-z_0}\right.\text{ are collinear}$

$\Leftrightarrow \exists \lambda \in \mathbb{R} , \left\{\matrix{x-x_0=\lambda(x_1-x_0)\\y-y_0=\lambda(y_1-y_0)\\z-z_0=\lambda(z_1-z_0)}\right.$

$\Leftrightarrow \exists \lambda \in \mathbb{R} , \frac{x-x_0}{x_1-x_0}=\frac{y-y_0}{y_1-y_0}=\frac{z-z_0}{z_1-z_0}=\lambda$

Answer 3

Note that the vector connecting the two points $(x_0, y_0, z_0)$ and $(x_1, y_1, z_1)$ within the line is $(x_1-x_0, y_1-y_0, z_1-z_0)$. Assume $(x,y,z)$ is a point in the line. Then the vector connecting the two points $(x, y, z)$ and $(x_0, y_0, z_0)$ is $(x-x_0, y-y_0, z-z_0)$.

Since the two vectors lie in the same line, they are proportional to each other, i.e.

$$(x-x_0, y-y_0, z-z_0)= c(x_1-x_0, y_1-y_0, z_1-z_0)$$

where $c$ is the proportional constant. Then, write the three component equations separately and equate them to the common constant $c$,

$$c=\dfrac{x - x_0}{x_1 - x_0} = \dfrac{y - y_0}{y_1 - y_0} = \dfrac{z - z_0}{z_1 - z_0}$$