What's the distance formula for this problem?

[watdahel]

Let's say I have 2 vertices with known coordinates, that form a line segment. I have another vertex; what's the best way to find out the distance of this vertex from the line segment? I guess I need the length of the line perpendicular to the line segment from this vertex.

 

[Firus]

You need to find some sort of relation between the other vertices and the one you are trying to find the distance to

 

[cjr22]

If your two original vertices are a and b, then the vector of the line running between them is a+n(b- a). IF your third vertex is c then the distance from c to a point on the line is the length of a+n(b- a) - c. IF you minimise this formula wrt n, then you'll get the minimum distance.

 

[notfred]

Assume A and B are your endpoints and C is your other point.

convert the coordinates to polar coordiantes, make point A the origin, and rotate B and C around the origin until B lies alogn the X axis of the previous rectangular coordinate system. Convert back to rectangular coordinates.

Your line segment is now along the X axis. If the X value of C is between the X value of A and B, then the distance is the Y coordinate of C. Otherwise, it's the distance to the nearest endpoint.

That's how I'd do it.

 

[Bulldog13]

Reminds me of the day of my Calc 3 final (assuming this isn t geometry or something )

"Screw this, I ll wake up at 5 in the morning and study all the different types of coord system and how to convert them. Midterm is at 9, will give me plenty of time"

:: hits the hay ::

:: Sleeping blissfully ::

Roommate : "Bulldog, bulldog...."

Me : "SSSHhhhhh, sleeping"

Roommate : "Ain t you got a final today ?"

Me : ::glances at clock :: 8:57 it reads

Me: OH SH*T, HOLY F*CK

Roommate : "Hows that studying going ?"

Me : "Die"

I ran straight to class in flip flops, shorts, t shirt and leather jacket. Nevermind it was poring.

Still swung a B+ too.

 

[notfred]

Here's C++ code that implements my solution.

 

[watdahel]

Thanks for that code. Did you wrote that up just for me? What's the order of input? Is it x2,y2,x1,y1,x3,y3.

Below is an equation I found that seems to show the same results as your algorithm when I enter the input in the order above.

d= |(x2-x1)(y1-y0) - (x1-x0)(y2-y1)| / sqrt( (x2-x1)² + (y2-y1)² )

 

[notfred]

Yeah, I wrote that in repsonse to this question. The order of input is in the comments above the loop that takes input.

Your algorithm is shorter, but I came up with mine myself