Math Is Fun Forum

Zenith

Member

Registered: 2008-03-24

Posts: 1

Calculate a point along a line

I have tried to solve this problem for a while now, and I always seem to get the wrong answer. I really need help.

Two points form a line. How do I find the the x and y coordinates of a point that is n units from one point going towards the other point along the line?

Diagram:

        n
  _________________________
A              B                          C

How do you find where B is? I turn the line into a triangle and use similar edges of it to find the answer. However, this do what I need; I want to use some sort of equation/rule to find the location of B.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Calculate a point along a line

Let's say that point A is (a,b) and point C is (c,d).

To get from A to C, you need to go (c-a) units to the right, and (d-b) units up.
So the equation of the line that contains points A and C is (a,b) + t(c-a,d-b), where t is some parameter. (This is because every point on the line can be made by starting at point A and then moving some distance parallel to the line AC.)

You can also find the length AC by Pythagoras: AC = √[(c-a)²+(d-b)²].
By introducing a new variable L and saying that t = L/AC, you get the equation of the line joining A and C to be (a,b) + L/AC(c-a,d-b). This time, L indicates the distance from point A (in the direction of C).

Hence, substituting your value of n into L will get your answer.

(Hopefully I've explained that well enough. Feel free to post again if I've been unclear.)

---

Why did the vector cross the road?
It wanted to be normal.

mikkel

Guest

Re: Calculate a point along a line

That was extremely helpful for my game project I am doing, so glad I found this explanation thank you!! :D

mathsfailure

Guest

Re: Calculate a point along a line

(x1,y1) + t(x2-x1,y2-y1)

AC = √[(x2-x1)²+(y2-y1)²]

t = L/AC

(x1,y1) + (L/√[(x2-x1)²+(y2-y1)²])(x2-x1,y2-y1)

I don't understand how to add to (x1,y1). (a,b) + 1 = ???
I don't understand how to extract a result after substituting L for a distance.

Let me try it anyway.
(x1,y1) = (10, 20)
(x2,y2) = (20, 40)
L = 5

x2-x1 = 10
y2-y1 = 20

AC = √(100 + 400) = 22.4
t = 5/22.4 = 0.223

(10, 20) + 0.223(10, 20)

I'm not sure what to do here but I'll assume I separate the x and y values like this:

x = 10 + (0.223 * 10)
y = 20 + (0.223 * 20)

Final answer:
(12.2, 24.5)

Bob

Administrator

Registered: 2010-06-20

Posts: 10,619

Re: Calculate a point along a line

hi mathsfailure

Welcome to the forum.

The distance between two points is given by your equation for AC

http://www.mathsisfun.com/algebra/dista … oints.html

Your calculation for AC is correct.

The formula

enables you to work out how far along the line to go.

Your calculation of t is correct.

Yes, it is correct to deal with the x coordinates and the y coordinates separately like that. 

So, ..... not so much of a failure after all. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob