如何获得与另一点的距离最小的线上的点的坐标 [英] How to get the coordinates of the point on a line that has the smallest distance from another point

问题描述

我现在正在为这个几何问题而苦苦挣扎.

i'm struggling with this geometry problem right now.

假设我们有一条由点A(x1，y1)和点B(x2，y2)定义的线我们还有一个点C(x3，y3).

Let's say we have a line defined by point A(x1,y1) and point B(x2,y2) We also have a point C(x3,y3).

用SWIFT编写的哪个函数可以给我距线最小距离的点的坐标(X，Y)?换句话说，线上的点是垂直线段与另一点之间的交点.

What function written in SWIFT could give me the coordinates (X,Y) of the point that has the smallest distance from the line ? In other words, the point on the line which is the intersection between a perpendicular segment and the other point.

func getCoordsOfPointsWithSmallestDistanceBetweenLineAndPoint(lineX1: Double, lineY1: Double, lineX2: Double, lineY2: Double, pointX3: Double, pointY3: Double) -> [Double] {
    // ???
    return [x,y]
}

推荐答案

从数学的角度来看，您可以:

In a mathematical point of view you can :

• 首先找到直线的方程式:

• first find the equation of the line :

    y1 = a1x1+b1 
    a1 = (y2-y1) / (x2-x1)
    b1 = y1-a1*x1

然后知道第二条线的斜率:

Then calculate the gradient of the second line knowing :

    a1 * a2 = -1 <-> 
    a2 = -1/a1

使用a2，您可以找到第二个方程式的b值:

with a2 you can find the value of b for the second equation :

    y3 = a2*x3 + b2 <-> 
    b2 = y3 - a2*x3

最后计算两条线的交点:

Finally calculate the intersection of the 2 lines :

    xi = (b2-b1) / (a1-a2)
    y = a1*xi + b1

然后将其迅速实现很简单:

Then it's quite straightforward to bring that to swift :

typealias Line = (gradient:CGFloat, intercept:CGFloat)

func getLineEquation(point1:CGPoint, point2:CGPoint) -> Line {
    guard point1.x != point2.x else {
        if(point1.y != point2.y)
        {
            print("Vertical line : x = \(point1.x)")
        }
        return (gradient: .nan, intercept: .nan)
    }
    let gradient = (point2.y - point1.y)/(point2.x-point1.x)
    let intercept = point1.y - gradient*point1.x
    return (gradient: gradient, intercept: intercept)
}

func getPerpendicularGradient(gradient:CGFloat) -> CGFloat
{
    guard gradient != 0 else {
        print("horizontal line, the perpendicilar line is vertical")
        return .nan
    }
    return -1/gradient
}

func getIntercept(forPoint point:CGPoint, withGradient gradient:CGFloat) -> CGFloat
{
    return point.y - gradient * point.x
}

func getIntersectionPoint(line1:Line, line2:Line)-> CGPoint
{
    guard line1.gradient != line2.gradient else {return CGPoint(x: CGFloat.nan, y: CGFloat.nan)}
    let x = (line2.intercept - line1.intercept)/(line1.gradient-line2.gradient)
    return CGPoint(x:x, y: line1.gradient*x + line1.intercept)
}

func getClosestIntersectionPoint(forLine line:Line, point:CGPoint) -> CGPoint
{
    let line2Gradient = getPerpendicularGradient(gradient:line.gradient)
    let line2 = (
        gradient: line2Gradient,
        intercept: getIntercept(forPoint: point, withGradient: line2Gradient))
    
    return getIntersectionPoint(line1:line, line2:line2)
}

func getClosestIntersectionPoint(forLinePoint1 linePoint1:CGPoint, linePoint2:CGPoint, point:CGPoint) -> CGPoint
{
    return getClosestIntersectionPoint(
        forLine:getLineEquation(point1: linePoint1, point2: linePoint2),
        point:point)
}

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