Help! Slope problem

[Valce]

How can I find the point of intersection of 2 line segments without using slopes?

Or if that's not possible... is there an easy way to code an exception for vertical lines when using?

Thanks!!

[Cervantes]

Assuming you know that the two lines segments touch, if the slope of one of your lines is vertical, you can solve for the point of intersection by subbing in the x value of the vertical line into the x in the equation for your second line.
In other words, you know the x value of the point of intersection is going to equal the x value of the vertical line. This is because any point on the vertical line has the same x value. So, you take that value and sub it into your second equation and solve.

[Valce]

Thanks!


Is there an easy way to implement this? I have right now:
(I didn't include the beginning of this thing since it's built into a rather long shell... but basically a, b, a1, and b1 are endpoints and there are arrays of points/lines)

code:

if b1.x not= a1.x and b.x not= a.x then                     slope1 := (b1.y - a1.y) / (b1.x - a1.x)                     intercept1 := a1.y - slope1 * a1.x                     slope := (b.y - a.y) / (b.x - a.x)                     intercept := a.y - slope * a.x                     if slope not= slope1 then                         pt (pointnum).x := (intercept - intercept1) / (slope1 - slope)                         pt (pointnum).y := slope * (pt (pointnum).x) + intercept                     else                         parallel := true                     end if                 else                     if b1.x = a1.x and b.x not= a.x then                         slope := (b.y - a.y) / (b.x - a.x)                         intercept := a.y - slope * a.x                         pt (pointnum).x := a1.x                         pt (pointnum).y := slope * a1.x + intercept                     elsif b.x = a.x and b1.x not= a1.x then                         slope1 := (b1.y - a1.y) / (b1.x - a1.x)                         intercept1 := a1.y - slope1 * a1.x                         pt (pointnum).x := a.x                         pt (pointnum).y := slope1 * a.x + intercept1                     else                         parallel := true                     end if                 end if                 %display the point                 if parallel = true then                     winID := Window.Open ("position:top;center,text:200;200")                     put "Point of intersection not available"                     put "Line number ", linenum, " and ", linenum1, " are parallel"                     put "Press any key to continue"                     ch := getchar                     Window.Close (winID)                 else                     selpt (0) := pointnum                     DisplayPoint (pt (pointnum), dp, selcolour (0))                     DisplayPointData (0, pointnum, pt (pointnum), cp)                     if pt (pointnum).x < -310 or pt (pointnum).x > cp.left or pt (pointnum).y < -310 or pt (pointnum).y > maxy then                         winID := Window.Open ("position:top;center,text:200;200")                         put "Point outside of screen. Intersection located at (", pt (pointnum).x, ", ", pt (pointnum).y, ")"                         put "Point stored as ", pointnum                         put "Press any key to continue"                         ch := getchar                         Window.Close (winID)                     end if                 end if                 parallel := false

And while it seems to work, it's also a tad long


P.S. Sorry the code sort of runs down a line... The box is too small

[MysticVegeta]

Make 2 2d arrays: line1(x, y) and line2(x, y)

code:

for s : 1..upper(line1) if line1(x, y) = line2(x, y) intersect:= true end if end for

point of intersection can be found using this. Maybe lol

[Cervantes]

That would require a ton of dots, and way too much processing time.
The idea is to find the point of intersection using math, not brute force. Math is always nicer than brute force.
Also, consider if your points were real numbers. You'd have to have such an insanely high number of points to actually get an answer. If you round things, you lose accuracy.
So math is more accurate and faster.
Sorry Valce, can't look over your code right now. Hopefully someone else has the time.