查找一个点是否在三角形内 [英] Finding whether a point is within a triangle

问题描述

我已经在这上面待了好几个小时，试着用不同的方法来看待每一个问题。也许我完全错了，但我觉得我的数学是正确的，但无论输入什么数字，我都会得到相同的输出。我的代码已关闭，我必须在午夜之前将其关闭。

它非常有趣：查找一个点是否在三角形代码中。 （适用于初学者）

$ $ p $ import java.util.Scanner;

public class PointsTriangle {

//检查输入的点是否在三角形内
//给定的三角形点是（0,0）（0,100）（ 200,0）
public static void main（String [] args）{
//从用户$ b $获取点（x，y）System.out.print（输入一个点的x-和y坐标：）;
扫描仪输入=新扫描仪（System.in）;
double x = input.nextDouble（）;
double y = input.nextDouble（）;

//找到给定点的三角形面积
double ABC =（（0 *（100-0）+ 0 *（0-0）+ 200 *（0-100）） /2.0）;
double PAB =（（x *（0-100）+ 0 *（100-y）+0 *（y-0））/2.0）;
double PBC =（（x *（100-0）+ 0 *（0-y）+ 200 *（y-100））/2.0）;
double PAC =（（x *（0-100）+ 0 *（100-y）+ 200 *（y-0））/2.0）;

boolean isInTriangle = PAB + PBC + PAC == ABC;

if（isInTriangle）
System.out.println（点在三角形中）;
else
System.out.println（该点不在三角形中）;
} //结束主
} //结束PointsTriangle


解决方案

如果您绘制图片，您可以看到该点必须满足简单的不等式（在特定线条右侧的下方/上方/）。无论是在边缘还是在外，我都会留给你：

  Y> 0（在X轴上方）
 X> 0（在Y轴的右边）
 X + 2 * Y < 200（在斜边以下）
  

在这三个地方写一条if语句， （（y> 0）&（x> 0）&&（x + 2 * y））（b>

  < 200））
 System.out.println（点在三角形中）; 
 else 
 System.out.println（该点不在三角形中）; 
  

I have been on this for hours, attempting different methods looking at just about every question. Perhaps I have it completely wrong, but I feel that I have my math of it correct, but no matter what numbers I input, I get the same output. My code is off somewhere and I have to turn it in by midnight.

It is the all so fun: Find if a point is within a triangle code. (for beginners)

import java.util.Scanner;

public class PointsTriangle {

    // checks if point entered is within the triangle
    //given points of triangle are (0,0) (0,100) (200,0)
    public static void main (String [] args) {
        //obtain point (x,y) from user
        System.out.print("Enter a point's x- and y-coordinates: ");
        Scanner input = new Scanner(System.in);
        double x = input.nextDouble();
        double y = input.nextDouble();

        //find area of triangle with given points
        double ABC = ((0*(100-0  )+0*(0  -0)+200*(0-100))/2.0);
        double PAB = ((x*(0  -100)+0*(100-y)+0  *(y-  0))/2.0);
        double PBC = ((x*(100-0  )+0*(0  -y)+200*(y-100))/2.0);
        double PAC = ((x*(0  -100)+0*(100-y)+200*(y-  0))/2.0);

        boolean isInTriangle = PAB + PBC + PAC == ABC;

        if (isInTriangle)
            System.out.println("The point is in the triangle");
        else
            System.out.println("The point is not in the triangle");
    }//end main
}//end PointsTriangle

解决方案

If you draw a picture, you can see the point has to satisfy simple inequalities (below / above / to the right of certain lines). Whether "on the edge" is in or out I will leave up to you:

Y > 0 (above the X axis)
X > 0 (to the right of the Y axis)
X + 2* Y < 200 (below the hypotenuse)

Write an if statement around these three and you're done:

if( (y > 0) && (x > 0) && (x + 2*y < 200) ) 
  System.out.println("The point is in the triangle");
else
  System.out.println("The point is not in the triangle");

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