CODEWITHSUNDEEP
c++
Mehar Ali
Mehar Ali

Reputation: 79

Find whether the triangle's origin lies inside the triangle or not

I am trying to find whether the given triangle has its origin inside or outside.

The code below always gives Origin is not inside given triangle. I have no idea why is it so and how to fix it.

float distance(int x1, int y1,int x2,int y2)
{
    int dis = (((x2-x1)*(x2-x1)+(y2-y1)*(y2-y1)));
    float dis1 = sqrt(dis);
    return dis1;
}

float AreaOfTriangle(float a, float b, float c)
{
    float s = (a+b+c)/2;
    float Area = sqrt(s*(s-a)*(s-b)*(s-c));
    cout << Area << endl;
    return Area;
}

int main()
{
    float dis1 = distance(-1,-1,0,1);
    float dis2 = distance(1,-1,-1,-1);
    float dis3 = distance(0,1,1,-1);

    float area = AreaOfTriangle(dis1,dis2,dis3);
    float dis4 = distance(0,0,-1,-1);
    float dis5 = distance(0,0,1,-1);

    float area1 = AreaOfTriangle(dis2,dis4,dis5);
    float dis6 = distance(0,1,0,0);

    float area2 = AreaOfTriangle(dis1,dis4,dis6);
    float area3 = AreaOfTriangle(dis3,dis5,dis6);
    float a = area1 + area2 + area3;

    cout << endl << a;

    if(area == a)
    {
        cout << "Origin is Inside Given Triangle";
    }
    else
    {
        cout << "Origin is not Inside Given Triangle";
    }

    return 0;
}

Upvotes: 2

Views: 756

Answers (2)

Jimmy
Jimmy

Reputation: 6171

I think you can do this

Consider points a,b and c (ax,ay),(bx,by) and (cx,cy)

define A = ax + ay, B = bx + by and C = cx + cy

Define z = y (B/A) + (1 -y) (C/A)

Calculate z for y=0 and y = 1. If either or both values are between 0 and 1 then the triangle contains the origin

Upvotes: 1

dBlisse
dBlisse

Reputation: 811

Given points a,b,c.

  • If a, b or c contain the origin then the triangle abc contains the origin.
  • If a, b and c are all to the left or to the right of the origin, then abc cannot contain the origin.
  • If a, b and c are all above or below the origin, then abc cannot contain the origin.
  • If a, b are to the left of the origin, and c is to the right of the origin, then if lines ac and bc cross the origin, then abc contains the origin. (also true for a, to the left and b, c to the right of origin)
  • If a, b are to the left of the origin, and c is to the right of the origin, then if line ac is above the origin and bc is below the origin, then abc contains the origin. (also true for a, to the left and b, c to the right of origin)

Of course, this is very general and can also be extrapolated to any point, and the first 3 steps are optimizations. There are also little kinks that'll need to be worked out in the implementation, but should hold true.

You'll need to figure out how to properly do floating point math though.

Upvotes: 0

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