↓Skip to main content

Point to Line Distance in 2D Plane

·492 words·3 mins

Suppose that we have a straight line formed by point B and C, and we have another point A. What is the distance from point A to line BC?

There are various ways to calculate this distance. Using vector calculus is one of the easiest ways to do it.

From elementary math, we know that distance from A to line BC is the distance between point A and D, where D is a point in line BC and line AD is perpendicular to line BC.

So the problem really boils down to finding the distance between between point A and D. That is, we need to find the value of following equations:

\[\begin{equation} \Vert \overrightarrow{AD} \Vert = \sqrt{(D_x - A_x)^2 + (D_y - A_y)^2} \end{equation}\]

Since point D is in line BC, so vector \(\overrightarrow{BD}\) and vector \(\overrightarrow{DC}\) is parallel to each other, i.e., \(\overrightarrow{BD} = t\cdot \overrightarrow{DC}\). We also know that \(\overrightarrow{AD}\) is perpendicular to \(\overrightarrow{BC}\). From these two conditions, we have the following two equations:

\[\begin{gather} \frac{D_x - B_x}{C_x - D_x} = \frac{D_y - B_y}{C_y - D_y} \label{eq2} \\ (D_x - A_x)(C_x - B_x) + (D_y - A_y)(C_y - B_y) = 0 \label{eq3} \end{gather}\]

Simplifying \(\eqref{eq2}\), we get:

\[\begin{equation} (D_x - B_x)(C_y - D_y) - (D_y - B_y)(C_x - D_x) = 0 \label{eq4} \end{equation}\]

Transforming \(\eqref{eq4}\) a bit, and we get:

\[\begin{equation} \begin{split} [(D_x - A_x) + (A_x - B_x)][(A_y - D_y) + (C_y - A_y)] -\\ [(D_y-A_y) + (A_y-B_y)][(A_x-D_x) + (C_x - A_x)] = 0 \end{split}\label{eq5} \end{equation}\]

Now, to simplify calculation, we use the following substitution:

\[\begin{align} D_x - A_x &= m \label{eq6} \\ D_y - A_y &= n \label{eq7} \end{align}\]

Substitute m and n into equation \(\eqref{eq3}\) and \(\eqref{eq5}\), and simplifying a bit, we now get:

\[\begin{gather} m(C_x-B_x) + n(C_y-B_y) = 0 \label{eq8} \\ m(C_y-B_y) - n(C_x - B_x) = (A_y - B_y)(C_x-A_x) - (A_x - B_x)(C_y-A_y) \label{eq9} \end{gather}\]

From equation \(\eqref{eq8}\) and \(\eqref{eq9}\), it it now easy to derive the value of \(m\) and \(n\):

\[\begin{align} m &= - \frac{(C_y - B_y)[A_y(B_x - C_x) - B_y(A_x - C_x) + C_y(A_x - B_x)]}{(C_x - B_x)^2 + (C_y - B_y)^2}\\ n &= \frac{(C_x - B_x)[A_y(B_x - C_x) - B_y(A_x - C_x) + C_y(A_x - B_x)]}{(C_x - B_x)^2 + (C_y - B_y)^2}\\ \end{align}\]

So the square sum of \(m\) and \(n\) is:

\[\begin{equation} m^2 + n^2 = \frac{[A_y(B_x - C_x) - B_y(A_x - C_x), + C_y(A_x - B_x)]^2}{(C_x - B_x)^2 + (C_y - B_y)^2} \end{equation}\]

The norm of vector \(\overrightarrow{AD}\) is:

\[\begin{equation} \Vert \overrightarrow{AD} \Vert = \frac{\lVert A_y(B_x - C_x) - B_y(A_x - C_x), + C_y(A_x - B_x)\rVert}{\sqrt{(C_x - B_x)^2 + (C_y - B_y)^2}} \end{equation}\]

In the above equation, the denominator is the norm of vector \(\overrightarrow{BC}\), the nominator is the norm of cross product between \(\overrightarrow{BC}\) and \(\overrightarrow{BA}\).

So the distance from point A to line BC can also be written as

\[\begin{equation} \lVert \overrightarrow{AD} \rVert = \frac{\lVert \overrightarrow{BC} \times \overrightarrow{BA} \rVert}{\lVert \overrightarrow{BC} \rVert} \end{equation}\]

References

  • https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line
  • https://stackoverflow.com/q/39840030/6064933
  • https://math.stackexchange.com/questions/330269/the-distance-from-a-point-to-a-line-segment