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I have the coordinates of 2 points A1 (x1,y1) and A2 (x2,y2) and a distance d. I need to find the coordinates of point A3 that is the distance d from point A2 on the linear graph defined by A1 and A2. How can i do that with JavaScript? similar to where the angle is known

I have the coordinates of 2 points A1 (x1,y1) and A2 (x2,y2) and a distance d. I need to find the coordinates of point A3 that is the distance d from point A2 on the linear graph defined by A1 and A2. How can i do that with JavaScript? similar to https://softwareengineering.stackexchange./questions/179389/find-the-new-coordinates-using-a-starting-point-a-distance-and-an-angle where the angle is known

• javascript
• line
• points
• calculation

Share Improve this question Follow edited Nov 24, 2017 at 10:18 user24957 asked Nov 24, 2017 at 9:46 user24957user24957 31533 silver badges1111 bronze badges

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2 Answers 2

Sorted by: Reset to default 9

var A1 = {
    x : 2,
    y : 2
};

var A2 = {
    x : 4,
    y : 4
};

// Distance
var  d= 2;

// Find Slope of the line
var slope = (A2.y-A1.y)/(A2.x-A1.x);

// Find angle of line
var theta = Math.atan(slope);

// the coordinates of the A3 Point
var A3x= A2.x + d * Math.cos(theta);
var A3y= A2.y + d * Math.sin(theta);

console.log(A3x);
console.log(A3y);

So you have to start from A1 and go to A2 direction by the distance between A1 and A2 + d
Assuming your point are objects of a Point class with x and y properties and a distance method you can do this:

function move_to(origin, direction, dist){
    let dx = direction.x - origin.x;
    let dy = direction.y - origin.y;
    let coef = dist / origin.distance(direction);

    let x = origin.x + dx * coef;
    let y = origin.y + dy *coef;
    return new Point(x, y)
}

move_to(A1, A2, A1.distance(A2) + d)

Here's a simple Point class implementation if you want:

class Point {

    constructor(x, y){
        this.x = x;
        this.y = y;
    }

    distance(point){
        return Math.sqrt((this.x - point.x) ** 2 + (this.y - point.y) ** 2)
    }

}

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