Distance between two parallel lines

We can define the distance between two parallel lines in the plane.

The problem

We have two parallel lines, p: −x +3y − 16 = 0 and q: −x + 3y − 4 = 0. How do we find the distance between them?

We can easily reduce the distance between such lines to the distance from a point to a line. All we need to do is choose a point on one of the lines and then work out the distance from this point to the other line.

On the line q we can choose the point A(8, 4). By substituting into the equation of q we check that it really is a point on this line:

$$\begin{array}{rcl}-x+3y-4& =& 0\\ -8+3\cdot 4-4& =& 0\\ -8+12-4& =& 0\\ 0& =& 0\end{array}$$

Now we can work out the distance from the point A to the line p. For this we have the formula

$$\begin{array}{rcl}d(A,p)& =& \frac{|a\cdot a_1+b\cdot a_2+c|}{\sqrt{a^2+b^2}}\end{array}$$

So we just substitute:

$$\begin{array}{rcl}d(A,p)& =& \frac{|a\cdot a_1+b\cdot a_2+c|}{\sqrt{a^2+b^2}}\\ d(A,p)& =& \frac{|-1\cdot 8+3\cdot 4-16|}{\sqrt{(-1{)}^2+3^2}}\\ & =& \frac{12}{\sqrt{10}}\approx 3.795\end{array}$$