直线上求点

问题描述：

以起点（x1，y1）和已知点（x2，y2）的直线上，求从起点开始，长度为r的点（所要求的点）的坐标

解决方案：

1.根据直线方程y=kx+b，可以求出k、b的值

2.根据圆的一般方程(x-a)^2+(y-b)^2=r^2,加上上面的公式，就可以计算出来了

实现代码：

public static short[] getStraightLinePoint(int x1, int y1, int x2, int y2,
			int radius) {
		short x3 = 0;
		short y3 = 0;
		float k = 0;
		float b = 0;

		if ((x2 - x1) == 0 && (y2 - y1 == 0)) {
			return null;
		}

		if ((x2 - x1) == 0) {
			if (y2 > y1) {
				x3 = (short) x1;
				y3 = (short) (y1 + radius);
				return new short[] { x3, y3 };
			} else {
				x3 = (short) x1;
				y3 = (short) (y1 - radius);
				return new short[] { x3, y3 };
			}
		}

		if ((y2 - y1 == 0)) {
			if (x2 > x1) {
				y3 = (short) y1;
				x3 = (short) (x1 + radius);
				return new short[] { x3, y3 };
			} else {
				y3 = (sho