### Instruction: Compute the distance between the parallel lines given by
\[\begin{pmatrix} 2 \\ -3 \end{pmatrix} + t \begin{pmatrix} 1 \\ -7 \end{pmatrix}\]and
\[\begin{pmatrix} 1 \\ -5 \end{pmatrix} + s \begin{pmatrix} 1 \\ -7 \end{pmatrix}.\].### Response: The distance between two parallel lines is the same as the length of the segment that is perpendicular to both lines and joins any two points on them.
I can choose any two points, but it might be easier to use the points where t = 0 and s = 0, which are the initial positions of the parametric equations.
These points are (2, -3) and (1, -5).
To find the segment that is perpendicular to both lines, I need to find a vector that is orthogonal to the direction vector of the lines, which is (1, -7).
One way to find such a vector is to use the dot product property, which says that two vectors are orthogonal if and only if their dot product is zero.
So I need to find a vector (a, b) such that (a, b) · (1, -7) = 0.
Expanding the dot product, I get a - 7b = 0.
There are infinitely many solutions for this equation, but I can choose any one that is convenient.
For example, I can choose a = 7 and b = 1, so the vector is (7, 1).
Now I have a vector that is perpendicular to both lines, and I can use it to find the length of the segment that joins the two points.
To do this, I can use the projection formula, which says that the length of the projection of a vector u onto a vector v is given by |u · v| / |v|.
In this case, u is the vector that goes from (2, -3) to (1, -5), which is (-1, -2), and v is the vector that is perpendicular to the lines, which is (7, 1).
So the length of the projection is |(-1, -2) · (7, 1)| / |(7, 1)|.
Computing the dot product, I get (-1, -2) · (7, 1) = -7 - 2 = -9.
Computing the magnitude of v, I get |(7, 1)| = √(49 + 1) = √50.
So the length of the projection is |-9| / √50 = 9 / √50.
This is the distance between the two parallel lines.