#include <iostream>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <cstdio>
#include <math.h>
 
#define pi 3.141592653589793
using namespace std;
/*已知两点(x1,y1)(x2,y2)求直线方程(ax+by+c=0) ：
解得：a=y1-y2,b=x2-x1,c=x1*y2-x2*y1;*/
//x=(c1-b1c1+b1c2)/(a1b1-a2b1-a1)
//y=-[a1*(c1-c1b1+b1c2)/(a1b1-a2b1-a1)+c1]/b1
int main(){
    int n,i=0;
    cin>>n;
double x1,y1,x2,y2,x3,y3,x4,y4;
cout<<"INTERSECTING LINES OUTPUT"<<endl;
	for (;i<n;i++)
    {
        cin>>x1>>y1>>x2>>y2>>x3>>y3>>x4>>y4;
        double a1=y1-y2,b1=x2-x1,c1=x1*y2-x2*y1;
        double a2=y3-y4,b2=x4-x3,c2=x3*y4-x4*y3;
       // cout<<"First line:"<<a1<<"x"<<"+"<<b1<<"y+"<<c1<<"=0"<<endl;
        //cout<<"Second line:"<<a2<<"x"<<"+"<<b2<<"y+"<<c2<<"=0"<<endl;
        //cout<<endl;
        if ((a1/a2)==(b1/b2)||(a1==0&&a2==0)||(b1==0&&b2==0))
        {
            if ((a1/a2)==(c1/c2)&&(b1/b2)==(c1/c2))
                cout<<"LINE"<<endl;//重合
            else if (c1==0&&c2==0)
                cout<<"LINE"<<endl;//重合
            else
                cout<<"NONE"<<endl;//平行
        }
        else
        {
           double x=(b2*c1-b1*c2)/(b1*a2-b2*a1);
           double y=(a2*c1-a1*c2)/(a1*b2-a2*b1);
 
           printf("POINT %.2lf %.2lf\n",x,y);
        }
        //cout<<endl;
    }
    cout<<"END OF OUTPUT"<<endl;
	return 0;
}
 

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10 
0 0 4 4 0 4 4 0 
5 0 7 6 1 0 2 3 
5 0 7 6 3 -6 4 -3 
2 0 2 27 1 5 18 5 
0 3 4 0 1 2 2 5 
0 0 1 1 0 0 1 1 
4 0 6 0 1 2 6 9 
1 0 5 0 8 5 4 5 
1 0 5 0 5 0 1 0 
0 1 0 5 0 5 0 1