Distance of two parallel lines
Kapitoly: Distance of a point from a line, Distance of a point from a plane, Distance of two straight lines
We can define a distance between two parallel lines in a plane.
Assignment
We have two parallel lines, p: −x +3y − 16 = 0 and q: −x + 3y − 4 = 0. How do we find the distance of these two lines?
We can easily reduce the distance of such lines to the distance of a point from the line. Just choose a point on one of the lines and then calculate the distance of that point from the other line.
We can choose a point on the line q A[8, 4] . By substituting q into the equation of the line, we can verify that it is indeed a point on that line:
\begin{eqnarray} -x + 3y - 4 &=& 0\\ -8 + 3\cdot4 - 4 &=& 0\ -8+12-4 &=& 0\ 0 &=& 0 \end{eqnarray}
Now we calculate the mentioned distance of the point A from the line p. For this we have the formula
\begin{eqnarray} v(A, p) &=& \frac{|a\cdot a_1+b \cdot a_2+c|}{\sqrt{a^2+b^2}} \end{eqnarray}
So we just add:
\begin{eqnarray} v(A, p) &=& \frac{|a\cdot a_1 + b \cdot a_2+c|}{\sqrt{a^2+b^2}}\\ v(A, p) &=& \frac{|-1\cdot 8 + 3 \cdot 4-16|}{\sqrt{(-1)^2+3^2}}\&& \frac{12}{\sqrt{10}}\approx 3,795 \end{eqnarray}