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VistasDetect interception with line
How can I make a algorithm that detects if a point (x, y) intercepts with a line. (x1, y1, x2, y2)?
I have already tried :
boolean onLine(float a, float b, float c, float d, float x, float y){
boolean answer = false;
float[] p1 = new float[] {a, b};
float[] p2 = new float[] {c, d};
float x_spacing = (p2[0] - p1[0]) / ((a+c)/2 + (b+d));
float y_spacing = (p2[1] - p1[1]) / ((a+c)/2 + (b+d));
List<float[]> line = new ArrayList();
float currentX = 0;
float currentY = 0;
while(currentX+a<c&¤tY+b<d){
currentX += x_spacing;
currentY += y_spacing;
line.add(new float[]{a+currentX, b+currentY});
}
for(int j = 0; j < line.size(); j++){
if(x > line.get(j)[0]-x_spacing && x < line.get(j)[0]+x_spacing && y > line.get(j)[1]-
y_spacing && y < line.get(j)[1]+y_spacing){
answer = true;
println("Hit line!");
break;
}
}
return answer;
}
This works sometimes, but is not always consistent.
I am putting this with a physics game, and I need this so the ball can roll down a line.
What are some ways I can improve it so that it works?.
EDIT: Thanks to Felix Castor I got it working. Here is the final Code:
boolean onLine(float x1, float y1, float x2, float y2, float xt, float yt,
float wid, float hit){
float Y = (y2 - y1)/(x2 - x1)* xt + y1 -(y2 - y1)/(x2 - x1) * x1;
boolean answer = false;
if(abs(Y - yt) < 5) answer = true;
if(abs(Y - yt-hit) < 5) answer = true;
if(abs(Y - yt-(hit/2)) < 5) answer = true;
if(abs(Y - yt+hit) < 5) answer = true;
if(abs(Y - yt+(hit/2)) < 5) answer = true;
return answer;
}
1 Respuestas
Responde la pregunta0
Using slope intercept form you can plug in your x and see if the y's are equal.
y = m*x + b
m = (y2 - y1)/(x2 - x1)
b = y1 - (y2 - y1)/(x2 - x1) * x1
So the equation becomes
Y = (y2 - y1)/(x2 - x1)* X + y1 -(y2 - y1)/(x2 - x1) * x1
given a point (xt, yt) you can plug in the xt into X and evaluate then compare the result to yt. If they are equal then the point is on the line.
if Y == yt given xt then the point is on the line.
You will need to handle the case where you have strictly horizontal lines as edge cases. Those will blow up the equation.
Edit: Conditions Changed
Since you are wanting to determine how far from the line a point is I would say the formula for the distance between a point and a line in cartesian space would be the way to go. See Distance from a point to a line section Line Defined By Two Points. The formula looks ugly but it's straight forward.
double numerator = Math.abs((y2 - y1) * xt - (x2 - x1) * yt + x2 * y1 - y2 * x1);
double denominator = Math.sqrt(Math.pow(y2 - y1, 2) + Math.pow(x2 - x1, 2));
double distance = numerator / denominator;
As before your test point is (xt, yt) and your line is defined by two points (x1, y1) and (x2, y2). Because distance is always >= 0 your test would be:
if( distance <= tolerance) return true
I think this is a better approach if you are interested in a tolerance.