Parallel lines and Vectors

Question

Compute the distance between the parallel lines given by $\binom{1}{4}+t\binom{4}{3}$ and $\binom{-5}{6}+s\binom{4}{3}$. I'm not even sure if those are parallel lines! How would I know that those are parallel lines?

Answer 1

Your lines are parallel because they are both translates of the the line $\ell = \left\{t \left(\begin{matrix} 4 \\ 3 \end{matrix} \right) : t \in \mathbb{R} \right\}$.

Imagine sliding the line $\ell$ to a new position without rotating it so that it now goes through $\left( \begin{matrix} 1 \\ 4 \end{matrix} \right)$. This new line $\ell_1$ will have the same slope as $\ell$.

We can similarly perform a slope-preserving slide of $\ell$ to a line $\ell_2$ passing through $\left( \begin{matrix} -5 \\ 6 \end{matrix} \right)$.

The lines $\ell_1,\ell_2,\ell$ will thus have the same slope and will be parallel. The first two are precisely the lines that you have parameterized in your original question.

Now the slope of $\ell$ is $m=\frac{3}{4}$.

The line $\ell_1$ with slope $m$ that goes through $\left(\begin{matrix} 1 \\ 4 \end{matrix} \right)$ will have the equation $y = \frac{3}{4}x + \frac{13}{4}$. Similarly, the line $\ell_2$ with slope $m$ that goes through $\left(\begin{matrix} -5 \\ 6 \end{matrix} \right)$ will have equation $y = \frac{3}{4}x + \frac{39}{4}$.

Now, the distance between the lines $\ell_1$ and $\ell_2$ will be the same as the distance from the point $a = \left(\begin{matrix} 1 \\ 4 \end{matrix} \right)$ to $\ell_2$ (note that $a$ falls on $\ell_1$). To find the distance from a point to a line, the following article will help: https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line

Answer 2

Because the coefficient vectors of the indeterminants are parallel, now they are even the same: $\left(\matrix{4\\3}\right)$.

Answer 3

Two parallel lines $L_1$ and $L_2$ are given:

$1) L_1 : (x,y) = (1,4) + t(4,3)$;

$ x = 1 + 4t; \, y = 4 + 3t$.

$2) L_2 : (x,y) = (-5,6) + s(4,3)$;

$ x = -5 +4s; \, y = 6 + 3s$.

Two straight lines in parameter form.

(Eliminating the parameter, I.e. solving one of the equations for t in eqs. 1) or for s in eqs. 2) and inserting into the equation of the other coordinate gives the more familiar form y = mx + c.)

Parallel lines: Every point on one line has the same distance to the other line.

Pick the point $(1,4)$ on $ L_1 (t=0).$

The squared distance from a point $(x_1,y_1)$ to a point $(x_2,y_2)$ is given by

$$d^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2.$$

Inserting $(x_1, y_1) = (1,4)$ and $(x_2, y_2) =(-5 + 4s,6 + 3s)$, a point on $L_2$, we get:

$d^2 = (1 - (-5 + 4s))^2 + (4 - (6 + 3s))^2 =$

$= (6 - 4s)^2 + (-2 - 3s)^2 = $

$= 36 - 48s + 16s^2 + 4 + 12s + 9s^2 $

$= 25s^2 - 36 s + 40 = $

$= 25(s - 18/25)^2 - (18)^2 / 25 + 40$.

The first term is $ \geq 0$, a square.

$Min:$

$d^2 = 40 - (18/5)^2$, at $\\s = 18/25$.

Hope this helps.