Compute the distance between the parallel lines given by

\[\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}\]
and
\[\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}.\]

one way: determine where a perpendicular line intersects each, and find the distance between the two points of intersection.

another way: determine the equation of one of the lines, then find the distance to it from a point on the other line, suing the formula

d = |Ax+By+C|/√(A^2+B^2)

Oh, parallel lines, huh? They must be like best friends who always stay close together. How cute!

Anyway, to find the distance between these two parallel lines, we need to find the distance between any two points, one on each line, that are closest to each other.

Let's call the points on the lines \(\mathbf{P}\) and \(\mathbf{Q}\), where \(\mathbf{P} = \begin{pmatrix} 1 \\ 4 \end{pmatrix}+ t \begin{pmatrix} 4 \\ 3 \end{pmatrix}\) and \(\mathbf{Q} = \begin{pmatrix} -5 \\ 6 \end{pmatrix}+ s \begin{pmatrix} 4 \\ 3 \end{pmatrix}\).

The distance between \(\mathbf{P}\) and \(\mathbf{Q}\) is given by the formula \(\|\mathbf{P} - \mathbf{Q}\|\).

Substituting the respective values into the formula, we have:

\[
\|\mathbf{P} - \mathbf{Q}\| = \left\| \left( \begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right) - \left( \begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right) \right\|.
\]

Now, let's simplify this expression:
\[
\|\mathbf{P} - \mathbf{Q}\| = \left\| \begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} - \begin{pmatrix} -5 \\ 6 \end{pmatrix} - s \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right\|.
\]

Okay, one step closer to finding the distance between these two lines!

Now, let's combine like terms:
\[
\|\mathbf{P} - \mathbf{Q}\| = \left\| \begin{pmatrix} 1 + (-5) \\ 4 + 6 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} - s \begin{pmatrix} 4 \\ 3 \end{pmatrix} \right\|.
\]

Adding the numbers together, we have:
\[
\|\mathbf{P} - \mathbf{Q}\| = \left\| \begin{pmatrix} -4 \\ 10 \end{pmatrix} + \begin{pmatrix} 4t \\ 3t \\
\end{pmatrix} - \begin{pmatrix} 4s \\ 3s \end{pmatrix} \right\|.
\]

Almost there! Let's simplify some more:
\[
\|\mathbf{P} - \mathbf{Q}\| = \left\| \begin{pmatrix} -4 + 4t - 4s \\ 10 + 3t - 3s \end{pmatrix} \right\|.
\]

Finally, the distance between the lines is given by the magnitude of this vector:
\[
\|\mathbf{P} - \mathbf{Q}\| = \sqrt{(-4 + 4t - 4s)^2 + (10 + 3t - 3s)^2}.
\]

There you have it! The distance between the parallel lines is \(\sqrt{(-4 + 4t - 4s)^2 + (10 + 3t - 3s)^2}\), where \(t\) and \(s\) can take any real values.

To find the distance between two parallel lines, we can take any point on one line and find the perpendicular distance to the other line. Let's choose a point on the first line, given by \[\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix}.\]

Let's find the perpendicular distance from this point to the second line.

The direction vector of the second line is $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$.

To find the perpendicular distance, we can find the projection of the vector connecting a point on the second line to the point on the first line onto the direction vector of the second line.

Let's choose a point on the second line, given by \[\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}.\]
The vector connecting the two points is \[\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} - \left(\begin{pmatrix} -5 \\ 6 \end{pmatrix} + s \begin{pmatrix} 4 \\ 3 \end{pmatrix}\right) = \begin{pmatrix} 6+4t \\ -2-3t \end{pmatrix} - \begin{pmatrix} -5+4s \\ 6+3s \end{pmatrix} = \begin{pmatrix} 11+4t-4s \\ -8-3t-3s \end{pmatrix}.\]
The projection of this vector onto the direction vector of the second line $\begin{pmatrix} 4 \\ 3 \end{pmatrix}$ is given by
\[\text{proj}_{\begin{pmatrix} 4 \\ 3 \end{pmatrix}} \begin{pmatrix} 11+4t-4s \\ -8-3t-3s \end{pmatrix} = \frac{\begin{pmatrix} 11+4t-4s \\ -8-3t-3s \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 3 \end{pmatrix}}{\begin{pmatrix} 4 \\ 3 \end{pmatrix} \cdot \begin{pmatrix} 4 \\ 3 \end{pmatrix}} \begin{pmatrix} 4 \\ 3 \end{pmatrix} = \frac{(11+4t-4s)(4) + (-8-3t-3s)(3)}{(4)(4) + (3)(3)} \begin{pmatrix} 4 \\ 3 \end{pmatrix}.\]

Now, we can find the perpendicular distance by finding the length of the vector connecting the chosen point on the first line to the projection:
\[\text{distance} = \left\|\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} - \left[\frac{(11+4t-4s)(4) + (-8-3t-3s)(3)}{(4)(4) + (3)(3)} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\right]\right\|.\]
Simplifying, we have
\[\text{distance} = \left\|\begin{pmatrix} 1 \\ 4 \end{pmatrix} + t \begin{pmatrix} 4 \\ 3 \end{pmatrix} - \frac{48+13t-25s}{25} \begin{pmatrix} 4 \\ 3 \end{pmatrix}\right\|.\]

Now, we can calculate the distance by expanding the expression inside the absolute value and simplifying.

To compute the distance between two parallel lines, we can use the following formula:

\[d = \frac{| \mathbf{r} \cdot \mathbf{n} |}{\| \mathbf{n} \|},\]

where \(\mathbf{r}\) is the vector connecting any point on one line to the other line, and \(\mathbf{n}\) is the unit normal vector to the direction vector of the lines.

In this case, the direction vector of both lines is \(\begin{pmatrix} 4 \\ 3 \end{pmatrix}\). To find the unit normal vector, we can take any vector perpendicular to the direction vector. One possible choice is \(\begin{pmatrix} -3 \\ 4 \end{pmatrix}\). We can then normalize this vector to get the unit normal vector:

\[\mathbf{n} = \frac{\begin{pmatrix} -3 \\ 4 \end{pmatrix}}{\|\begin{pmatrix} -3 \\ 4 \end{pmatrix}\|}.\]

Now, let's find a vector \(\mathbf{r}\) connecting a point on one line to the other line. We can take a point on the second line, say \(\begin{pmatrix} -5 \\ 6 \end{pmatrix}\), and subtract a point on the first line, say \(\begin{pmatrix} 1 \\ 4 \end{pmatrix}\), to get:

\[\mathbf{r} = \begin{pmatrix} -5 \\ 6 \end{pmatrix} - \begin{pmatrix} 1 \\ 4 \end{pmatrix} = \begin{pmatrix} -6 \\ 2 \end{pmatrix}.\]

Now, we can substitute these values into the formula for the distance:

\[d = \frac{| \begin{pmatrix} -6 \\ 2 \end{pmatrix} \cdot \left(\frac{\begin{pmatrix} -3 \\ 4 \end{pmatrix}}{\|\begin{pmatrix} -3 \\ 4 \end{pmatrix}\|}\right) |}{\| \frac{\begin{pmatrix} -3 \\ 4 \end{pmatrix}}{\|\begin{pmatrix} -3 \\ 4 \end{pmatrix}\|} \|}.\]

Evaluating this expression will give us the distance between the parallel lines.