Find the closest distance between a point and a segment defined by two points by coordinates

geometry

Let's assume $P_1=(x_1, y_1)$ and $P_2=(x_2, y_2)$ and $P_3=(x_3, y_3)$.

How to find the closest distance between $P_3$ and the line segment between $P_1$ and $P_2$?

I tried using the formula $\frac{area(P_1, P_2, P_3)}{distance(P_1, P_2)}$:
$$\operatorname{distance}(P_1, P_2, P_3) = \frac{|(x_2-x_1)(y_1-y_3)-(x_1-x_3)(y_2-y_1)|}{\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$$

Sadly it seems like it only works for infinite lines, not for line segments like in my case.

Best Answer

there is three area to take in count:

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For $P1$ the distance is $|P1-B|$
For $P3$ the distance is $|P3-A|$

to summarise, it juste the distance between two point when the point P is not between the two perpendicular line passing trough A and B.

the hardest part is P2 :

since it between the two point, the distance to the segment is the distance to the point $C$ on the segment $AB$. We can know where the point C is located with a parameter $h$ (the doted line) because of how a segment is parameterised.

$$(A,B) = (hx_B + (1-h)x_A,hy_B + (1-h)y_A)$$

if C is on B then $h=1$ and $h=0$ if C is on A, in the exemple h might be equal to 0,75 or 0,8.

To find h we need to do :

$$h = \frac{<P_2-A, B-A>}{|B-A|^2}$$

this is a projection of the length of the vector $\overrightarrow{P_2A}$ on to $\overrightarrow{BA}$ we use a dot product for that, then devide it to the length of $\overrightarrow{BA}$ then normalise it ench the power of 2.

then to find the distance $|P2-C|$ we do :

$$|P_2 - (A +h*(B-A))|$$

to finish we can compute the 3 expression at the same time wich lead us to :

$$h = min(1,max(0,\frac{<P_2-A, B-A>}{|B-A|^2}))$$ $$d = |P - A - h*(B-A)|$$

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