[计算几何]求两个直线交点坐标

上一篇文章已经讲了若一直线过 $A(X_1,Y_1)$,$B(X_2,Y_2)$ 两点的解析式求法

其$ax+by+c=0$形式的解析式 $a=(Y_1-Y_2),b=(X_2-X_1),c=(X_1\ast Y_2-Y_1\ast X_2)$

我们能求得 两个直线解析式分别为$a_{1}x+b_{1}y+c_1=0$和$a_{2}x+b_{2}y+c_2=0$

• 求x坐标

$y=\frac{-a_{1}x-c_1}{b_1}$,$y=\frac{-a_{2}x-c_2}{b_2}$

$\frac{-a_{1}x-c_1}{b_1}=\frac{-a_{2}x-c_2}{b_2}$

$(-a_{1}x-c_1)b_2=(-a_{2}x-c_2)b_1$

$-a_1{b}_{2}x-c_1{b}_2=-a_2{b}_{1}x-c_2{b}_1$

$a_2{b}_{1}x-a_1{b}_{2}x=c_1{b}_2-c_2{b}_1$

$(a_2{b}_1-a_1{b}_2)x=c_1{b}_2-c_2{b}_1$

$x=\frac{c_1{b}_2-c_2{b}_1}{a_2{b}_1-a_1{b}_2}$

• 求y坐标

$x=\frac{-b_{1}y-c_1}{a_1}$,$x=\frac{-b_{2}y-c_2}{a_2}$

$\frac{-b_{1}y-c_1}{a_1}=\frac{-b_{2}y-c_2}{a_2}$

$(-b_{1}y-c_1)a_2=(-b_{2}y-c_2)a_1$

$-b_1{a}_{2}y-c_1{a}_2=-b_2{a}_{1}y-c_2{a}_1$

$b_2{a}_{1}y-b_1{a}_{2}y=c_1{a}_2-c_2{a}_1$

$(b_2{a}_1-b_1{a}_2)y=c_1{a}_2-c_2{a}_1$

$y=\frac{c_1{a}_2-c_2{a}_1}{b_2{a}_1-b_1{a}_2}$

c++ 代码

Point GSP(Line t1,Line t2){
    Point t;
    double a1=t1.a.Y-t1.b.Y,b1=t1.b.X-t1.a.X,c1=t1.a.X*t1.b.Y-t1.b.X*t1.a.Y;
    double a2=t2.a.Y-t2.b.Y,b2=t2.b.X-t2.a.X,c2=t2.a.X*t2.b.Y-t2.b.X*t2.a.Y;
    t.X=(c1*b2-b1*c2)/(b1*a2-a1*b2);t.Y=(c1*a2-a1*c2)/(a1*b2-b1*a2);
    return t;
}