OT: 2D/3D math

[narf poit chez BOOM]

Anyone know how to determine the closest point on a line to a point? Rapidly running out of ideas...

Thanks.

[Aiken]

crossing of perpendicular from this point with a line itself?

[Slick]

yeah for 2D:

line: Ax + By + C = 0
Point: (x1, y1)

[saving you the gory details]

Code:

---

                                      |Ax1 + By1 + C|

distance between point and line: D = -----------------

                                     square root (A^2 + B^2)

---

notation: | | is absoute value, ^2 is raised to power 2


For 3D, it takes a little more...


Slick.

[TerranC]

Mathematics! Formulas! Variables! Discussion!

Run away! Run away! Math is taking over the world!

[narf poit chez BOOM]

Code:

---

                                      |Ax1 + By1 + C|

distance between point and line: D = -----------------

                                     square root (A^2 + B^2)

---

So...
Code:

---

                                      |A*x1 + B*y1 + C|

distance between point and line: D = -----------------

                                     square root (A^2 + B^2)

---

??

In that case, what are A, B and C?

[Slick]

use some algebra to transform the equation of the line to:

line: Ax + By + C = 0

then use whatever you came up with for A, B & C in the formula I gave to find the distance between the point and the line.

Slick.

[narf poit chez BOOM]

Or, you know, I could do that thing I almost never think of and explain myself...

I'm programming. I have the start XY of the line, the end XY of the line and the XY of the point. And I have no idea how to teach a computer algebra derivation.

[Aiken]

If P1(x1,y1) - startpoint
and P2(x2,y2) - endpoint
Then equation of the line will be:
x*(y2-y1) - y*(x2-x1) - x1*y2 + x2*y1 = 0

so A = y2-y1; B = -(x2-x1); C = -(x1*y2 - x2*y1).

But how to find coordinates of the point in the cross of line and perpendicular?