已知两点坐标，求直线方程、距离其中一点距离为L的某点

private void pointXY() {
Point curPoint = new Point(20, 30);// 当前坐标
Point nextPoint = new Point(35, 42);// 下个点坐标
double distance = Math.sqrt(Math.pow(curPoint.x - nextPoint.x, 2)
+ Math.pow(curPoint.y - nextPoint.y, 2));// 两点的坐标距离
double lenthUnit = distance / 5;// 单位长度
// 第一步：求得直线方程相关参数y=kx+b
double k = (curPoint.y - nextPoint.y) * 1.0
/ (curPoint.x - nextPoint.x);// 坐标直线斜率k
double b = curPoint.y - k * curPoint.x;// 坐标直线b
// 第二步：求得在直线y=kx+b上，距离当前坐标距离为L的某点
// 一元二次方程Ax^2+Bx+C=0中,
// 一元二次方程求根公式：
// 两根x1,x2= [-B±√(B^2-4AC)]/2A
// ①(y-y0)^2+(x-x0)^2=L^2;
// ②y=kx+b;
// 式中x,y即为根据以上lenthUnit单位长度(这里就是距离L)对应点的坐标
// 由①②表达式得到:(k^2+1)x^2+2[(b-y0)k-x0]x+[(b-y0)^2+x0^2-L^2]=0
double A = Math.pow(k, 2) + 1;// A=k^2+1;
double B = 2 * ((b - curPoint.y) * k - curPoint.x);// B=2[(b-y0)k-x0];
int m = 1;
double L = m * lenthUnit;
// C=(b-y0)^2+x0^2-L^2
double C = Math.pow(b - curPoint.y, 2) + Math.pow(curPoint.x, 2)
- Math.pow(L, 2);
// 两根x1,x2= [-B±√(B^2-4AC)]/2A
double x1 = (-B + Math.sqrt(Math.pow(B, 2) - 4 * A * C)) / (2 * A);
double x2 = (-B - Math.sqrt(Math.pow(B, 2) - 4 * A * C)) / (2 * A);
double x = 0;// 最后确定是在已知两点之间的某点
if (x1 == x2) {
x = x1;
} else if (curPoint.x <= x1 && x1 <= nextPoint.x || nextPoint.x <= x1
&& x1 <= curPoint.x) {
x = x1;
} else if (curPoint.x <= x2 && x2 <= nextPoint.x || nextPoint.x <= x2
&& x2 <= curPoint.x) {
x = x2;
}
double y = k * x + b;
Point mPoint = new Point((int) x, (int) y);
}