Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 4619792
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 22, 20262026-05-22T02:28:20+00:00 2026-05-22T02:28:20+00:00

Suppose I have two points representing a line A, such as: var A =

  • 0

Suppose I have two points representing a line A, such as:

var A = [ { x: 385, y: 380 }, { x: 420, y: 400 }]

And I have two other points B and C such as:

var B = { x: 385, y: 420 }
var C = { x: 405, y: 423 }

How would I determine if B and C are both on the same side of line A? To add a little context, I am trying to do hit testing for a hexagon, where B is the center point of the hexagon, C is the current mouse position and A is each line of the hexagon. All of these points are essentially pixel coordinates where 0,0 is the upper left hand corner.

I don’t need this to be blazing fast I am just trying to create the simplest possible hexagon hit test algorithm I can. My theory is that if I can determine that C is on the same side of each line of the hexagon as B then the hit test is successful. I have read several mathematical algorithms for doing this but they always seem to be in a different type of coordinate system and I’m struggling to translate it into something usable in javascript.

EDIT: here is my actual Hexagon function given the answer below. The answer to this question is in the update function.

var TILE_WIDTH = 70
var TILE_HEIGHT = 80

function Hexagon(x, y) {
    var normalColor = 'rgb(207, 226, 243)'
    var hilightColor = 'rgb(204, 204, 204)'
    var currentColor = normalColor

    var coords = new TileCoordinates(x, y)
    var points = [
        { x: coords.x, y: coords.y - TILE_HEIGHT / 2 },
        { x: coords.x + TILE_WIDTH / 2, y: coords.y - TILE_HEIGHT / 4 },
        { x: coords.x + TILE_WIDTH / 2, y: coords.y + TILE_HEIGHT / 4 },
        { x: coords.x, y: coords.y + TILE_HEIGHT / 2 },
        { x: coords.x - TILE_WIDTH / 2, y: coords.y + TILE_HEIGHT / 4 },
        { x: coords.x - TILE_WIDTH / 2, y: coords.y - TILE_HEIGHT / 4 },
    ]

    var sides = [
        [points[0], points[1]],
        [points[1], points[2]],
        [points[2], points[3]],
        [points[3], points[4]],
        [points[4], points[5]],
        [points[5], points[0]]
    ]

    this.update = function (totalTime, updateTime) {

        var B = coords
        var C = Mouse.state
        var inside = C != null
        if (inside) {
            for (i in sides) {
                var A = sides[i]
                var w = { y: A[1].x - A[0].x, x: -(A[1].y - A[0].y) }
                var P = A[1]

                inside = ((B.x - P.x) * w.x + (B.y - P.y) * w.y) * ((C.x - P.x) * w.x + (C.y - P.y) * w.y) > 0
                if (!inside) break
            }
        }

        if (inside)
            currentColor = hilightColor
        else
            currentColor = normalColor
    }

    this.draw = function (ctx) {
        ctx.fillStyle = currentColor
        ctx.strokeStyle = 'rgb(11, 83, 148)'
        ctx.beginPath()
        ctx.moveTo(points[0].x, points[0].y)
        ctx.lineTo(points[1].x, points[1].y)
        ctx.lineTo(points[2].x, points[2].y)
        ctx.lineTo(points[3].x, points[3].y)
        ctx.lineTo(points[4].x, points[4].y)
        ctx.lineTo(points[5].x, points[5].y)
        ctx.lineTo(points[0].x, points[0].y)
        ctx.fill()
        ctx.stroke()

        ctx.fillStyle = '#000'
        var text = coords.pos_x + ',' + coords.pos_y
        var measure = ctx.measureText(text)
        ctx.fillText(text, coords.x - measure.width / 2, coords.y + 12 + (TILE_HEIGHT / 4))
    }
}

// this is in a separate function because other objects that render into the hex
// need the pixel coordinates of the tile also
function TileCoordinates(x, y) {
    this.pos_x = x
    this.pos_y = y
    this.x = x * TILE_WIDTH + ((y + 1) * TILE_WIDTH / 2)
    this.y = (y + 1) * (3 / 4 * TILE_HEIGHT)
}

To determine same-sidedness I multiplied the results of B and C and if the result is > 0 then they are either both positive or both negative. I’m rendering and updating the hexagon into a canvas on a loop using setInterval.

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-22T02:28:21+00:00Added an answer on May 22, 2026 at 2:28 am

    The line representing A is described by the vector v = { x: 420 - 385, y: 400 - 380 } = { x: 35, y: 20 } and the starting point P = { x: 385, y: 380 }. Given a vector (x, y) in 2d, the vector (y, -x) is always at right angles to it. So the vector w = { x: 20, y: -35 } is at right angles to v. Linear algebra tells us that the sign of (B - P) dot w tells us which side of the line you are on where dot is the standard dot product. (The line itself is at zero.)

    Therefore in your example the calculation we need to do is this:

    For B:
    (B - P) dot w
      = { x: 385 - 385, y: 420 - 380 } dot { x: 20, y: -35 }
      = { x: 0, y: 40} dot { x: 20, y: -35 }
      = (0 * 20) + (40 * (-35))
      = -1400
    
    For C:
    (C - P dot w
      = { x: 405 - 385, y: 423 - 380 } dot { x: 20, y: -35 }
      = { x: 20, y: 43} dot { x: 20, y: -35 }
      = (20 * 20) + (43 * (-35))
      = -1105
    

    Since the sign is the same, they are on the same side.

    In fact we can say more. If you are at the starting point of A and are facing the ending point, both points will be on your left hand side. (The left side will be negative, the right side will be positive.)

    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

Suppose I have two applications written in C#. The first is a third party
Suppose you have two models, User and City, joined by a third model CityPermission:
Suppose I have two classes with the same interface: interface ISomeInterface { int foo{get;
Suppose you have two seperate ASP.NET Web Application projects that both need to use
Suppose I have two branches of a project IMClient-MacOS and IMClient-Windows, and their code
Suppose that you have two huge files (several GB) that you want to concatenate
Suppose I have the following two strings containing regular expressions. How do I coalesce
Suppose I have a database table with two fields, foo and bar. Neither of
Suppose I have a tags table with two columns: tagid and contentid . Each
Suppose I have a dataset with those two immortal tables: Employee & Order Emp

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.