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I am working on a collision detection problem.

I wish to measure the distance from a point to a line segment.

i.e. https://en.wikipedia.org/wiki/Distan..._by_two_points

I need to know the point on the line segment that the point (not on the line) is closest to.

i.e.

say the line segment is defined by

x1,y1 = 10,10

and

x2,y2 = 15,20

and there is a point at 16,16

how far is the shortest distance from the point to the line?

and where on the line is this shortest distance?

kind regards W

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  • $\begingroup$ do you know the Heesian normal form of a straight line? $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '15 at 18:12
  • $\begingroup$ no; I am not a mathematician (sadly). I have a practical problem in solid mechanics that I wish to model. That is, the distance from a point to a line, and the point on that line where the distance is shortest. $\endgroup$ – William White Oct 23 '15 at 23:59
  • $\begingroup$ I've managed to work this out. $\endgroup$ – William White Oct 25 '15 at 23:39
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As you say, we have 3 points, $A(10,10)$, $B(15,20)$, and $C(16,16)$.

The distance formula says that the distance between $A(x_1,y_1)$ and $B(x_2,y_2)$ is $$\sqrt {(x_2-x_1)^2+(y_2-y_1)^2}$$

Here, the distance $AB$ is $5\sqrt5$

Similarly, $AC$ is $6\sqrt2$

And $BC$ is $\sqrt{17}$

Since the perpendicular is the smallest distance, let a point $D(x,y)$ on $AB$ such that $AB\underline{|} CD$, we can form 2 right triangles $\delta ADC$ and $\delta BDC$

We already know that $AB$ is $5\sqrt5$

Let $CD=x$

Applying the Pythagorean theorem:-

$AD=\sqrt{72-x^2}$

And

$BD=\sqrt{17-x^2}$

Since $AB=AD+BD$

Therefore

$$5\sqrt5=\sqrt{72-x^2}+\sqrt{17-x^2}$$

Solving, $x={6\over\sqrt5}$

Therefore, here the smallest distance is $6\over\sqrt5$

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Let the line is given by two distinct points $A_1=(x_1,y_1)$ and $A_2=(x_2,y_2)$. Let the point you are studying be $B=(X,Y)$.

The points of the segment can be parametrised by $t\in[0,1]$ as $$((1-t) x_1 + t x_2, (1-t) y_1 + t y_2),$$ and the square of the distance to them is therefore $$ d(t) = ((1-t) x_1 + t x_2 -X)^2 + ((1-t) y_1 + t y_2 -Y)^2 .$$

You need to minimise this function (it is a convex function, which is great) on a convex compact set $[0,1]$, which guarantees that the solution exists.

After expanding all terms in $d(t) $ you can rewrite it as

$$d(t) = a_2 t^2 + a_1 t + a_0, \quad t\in[0,1]$$

After that you have several cases to study, but nothing too complex. If you have troubles here, ask in comments.

Another approach would be to study the vectors $\vec x_1 = \vec{BA_1}$, $\vec x_2 = \vec{BA_2}$, and $\vec z = \vec{A_1A_2}$ and the triangle $A_1A_2B$. After loking carefully at the picture, you can get the following result.

The distance from the line to the point is $\frac{|\vec z \times \vec x_1|}{\|\vec z\|}$, where $\times$ stands for vector product. Let also $\cdot$ be a scalar product.

If $$(\vec x_1\cdot\vec z)(\vec x_2\cdot\vec z)<0,$$ then the distance you want is $\frac{|\vec z \times \vec x_1|}{\|\vec z\|}$

otherwise, the distance is $\min (\|\vec x_1\|,\|\vec x_2\|)$

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