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Let's say you have two points, $(x_0, y_0)$ and $(x_1, y_1)$.

The gradient of the line between them is:

$$m = (y_1 - y_0)/(x_1 - x_0)$$

And therefore the equation of the line between them is:

$$y = m (x - x_0) + y_0$$

Now, since I want another point along this line, but a distance $d$ away from $(x_0, y_0)$, I will get an equation of a circle with radius $d$ with a center $(x_0, y_0)$ then find the point of intersection between the circle equation and the line equation.

Circle Equation w/ radius $d$:

$$(x - x_0)^2 + (y - y_0)^2 = d^2$$

Now, if I replace $y$ in the circle equation with $m(x - x_0) + y_0$ I get:

$$(x - x_0)^2 + m^2(x - x_0)^2 = d^2$$

I factor is out and simplify it and I get:

$$x = x_0 \pm d/ \sqrt{1 + m^2}$$

However, upon testing this equation out it seems that it does not work! Is there an obvious error that I have made in my theoretical side or have I just been fluffing up my calculations?

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    $\begingroup$ Looks about right to me. In particular, it gives the right result for reasonable values of $m = 0$, $m = 1$, $m = \infty$. Maybe there is a bug in your implementation. $\endgroup$ – Rahul Jul 27 '12 at 15:28
  • $\begingroup$ Thanks for the quick reply Rahul, i've been trying to programme this and you are right! it was an error in my implementation. Thank you for taking the time to read my question! $\endgroup$ – Kel196 Jul 27 '12 at 15:34
  • $\begingroup$ See formula 14 here. $\endgroup$ – J. M. is a poor mathematician Jul 27 '12 at 15:49
  • $\begingroup$ @enzotib: It is considered impolite in this site to remove or add "thank you" comments. $\endgroup$ – Asaf Karagila♦ Jul 27 '12 at 18:55
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    $\begingroup$ @AsafKaragila: sorry, I didn't know, in other SE sites it is considered superfluous to have such comments. I will take it into account for the future. $\endgroup$ – enzotib Jul 27 '12 at 18:57
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Another way, using vectors:

Let $\mathbf v = (x_1,y_1)-(x_0,y_0)$. Normalize this to $\mathbf u = \frac{\mathbf v}{||\mathbf v||}$.

The point along your line at a distance $d$ from $(x_0,y_0)$ is then $(x_0,y_0)+d\mathbf u$, if you want it in the direction of $(x_1,y_1)$, or $(x_0,y_0)-d\mathbf u$, if you want it in the opposite direction. One advantage of doing the calculation this way is that you won't run into a problem with division by zero in the case that $x_0 = x_1$.

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  • $\begingroup$ Thanks Theophile, absolutely ace method. I didn't even think about vectors! $\endgroup$ – Kel196 Jul 27 '12 at 16:03
  • $\begingroup$ Can you elaborate a little bit more on this solution or point me to a link so that I can understand it a little more? A program written would be awesome as well. $\endgroup$ – Dewey Sep 13 '14 at 22:08
  • $\begingroup$ @Dewey Which part are you having trouble understanding? $\endgroup$ – Théophile Sep 14 '14 at 13:07
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    $\begingroup$ @Dewey The length of a vector $\mathbf v = (v_1,v_2)$ is defined as $||\mathbf v|| = \sqrt{v_1^2 + v_2^2}$. The vector $\mathbf v \over ||\mathbf v||$, that is, $\Big(\dfrac{v_1}{\sqrt{v_1^2 + v_2^2}}, \dfrac{v_2}{\sqrt{v_1^2 + v_2^2}}\Big)$, points in the same direction as $\mathbf v$ and has unit length. For example, if $\mathbf v = (3,4)$, then $\mathbf u = ({3 \over 5}, {4 \over 5})$. $\endgroup$ – Théophile Sep 20 '14 at 23:00
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    $\begingroup$ @jpwrunyan Ah, I see! Okay, first, $||\bf v||$ refers to the length of the vector $\bf v$, which you can calculate using Pythagoras's theorem. In other words, if ${\bf v}=(x,y)$, then its length is $\sqrt{x^2+y^2}$. Next, about $v_1^2$ and $v_2^2$: these are easier than you think! They're just a compact way of writing $(v_1)^2$ and $(v_2)^2$, so in fact they aren't related to matrix notation (although that was a good guess). So if ${\bf v}=(x,y)$, then $||{\bf v}||=\sqrt{x^2+y^2}$; this is exactly the same as my previous comment but with differently named variables. Does that make more sense? $\endgroup$ – Théophile May 31 at 21:48
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Let me explain the answer in a simple way.

Start point - $(x_0, y_0)$

End point - $(x_1, y_1)$

We need to find a point $(x_t, y_t)$ at a distance $d_t$ from start point towards end point.

Point on a line at a distance

The distance between Start and End point is given by $d = \sqrt{(x_1-x_0)^2+(y_1-y_0)^2}$

Let the ratio of distances, $t=d_t/d$

Then the point $(x_t, y_t) =(((1-t)x_0+tx_1), ((1-t)y_0+ty_1))$

When $0<t<1$, the point is on the line.

When $t<0$, the point is outside the line near to $(x_0,y_0)$.

When $t>1$, the point is outside the line near to $(x_1,y_1)$.

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    $\begingroup$ Shouldn't t = d / dt? $\endgroup$ – skibulk Apr 21 '16 at 1:54
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    $\begingroup$ @skibulk nope it's t = dt / t when I applied this to my code I followed your comment. But something was bugging out, turns out t = dt / t is correct. $\endgroup$ – Marlon Aug 9 '16 at 9:40
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    $\begingroup$ @Marlon Thank you for confirming it. $\endgroup$ – Sen Jacob Aug 9 '16 at 15:00
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    $\begingroup$ Oh yes, you're correct. For some reason it I was thinking dt was "distance total" of the segment, which is be the divisor. Bit I guess it's supposed to be "distance time". Anyway, it's correct according to the illustration. $\endgroup$ – skibulk Aug 9 '16 at 16:17
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    $\begingroup$ If you only want to extend the line by a percentage, you don't need to calculate the distance, where p is a decimal percentage: dt = d * p, t = (d * p) / d, thus t = p $\endgroup$ – Kyle Apr 11 at 17:23
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You can very easily find it with trigonometry!!

Let's say Xa and Xb are the two points of your line, and D is the distance between them. And you are looking to find Xc which is D2 away from Xa (as the diagram bellow):

enter image description here

You can easily find D:

euclidean distance between Xa and Xb

The formulas that you can find Xa, Xb, Xc, D and D2 are:

enter image description here

But SINa-b and SINa-c share the same the same corner, so they are equal:

enter image description here

Since you know the distance (D2) between Xa and Xc that you are looking for, you can easily solve the following:

enter image description here

In conclusion by solving the formula for D and the last one you are done. (You need one for the Y as well, just replace in the last one, X with Y )

Hope it helps!!

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  • $\begingroup$ thanks this was really helpful $\endgroup$ – Vincent Tang Nov 29 '18 at 4:39
  • $\begingroup$ sweet solution! $\endgroup$ – Peter Pohlmann Jul 17 at 19:41
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The easy way in rectangular coordinate systems is to use the vector formula
P = d(B - A) + A
where
A is the starting point (x0, y0) of the line segment
B is the end point (x1, y1)
d is the distance from starting point A to the desired collinear point
P is the desired collinear point

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  • $\begingroup$ I have compared and tested this solution with the answer from @SenJacob above and this is NOT correct. $\endgroup$ – Minh Nguyen Jul 13 '17 at 13:21
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    $\begingroup$ This is the same as Théophile's answer, but it omits to mention that you need to normalize (B-A) $\endgroup$ – Gordon Jul 17 '17 at 1:40
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I think you need to check $x_0 > x_1$ when you try to calculate $x$ (last equation in your calculation) then you determine it will be $(+)$ or $(-)$ in your equation.

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  • $\begingroup$ Thank you for your contribution. This site supports basic TeX syntax, which allows formulas to be nicely typeset as $x_0>x_1$, for example. There is a short TeX tutorial. You may want to try using TeX by editing your answer (the link edit is under your post). Welcome to Math.SE! $\endgroup$ – user53153 Jan 1 '13 at 5:47
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There is an other way to solve this problem.

The function $\rho:\mathbb R \to \mathbb R^2$ defined by

$$\rho(t) = (1-t)(x_0,y_0) + t(x_1,y_1)$$

describes the line through the points $(x_0,y_0)$ and $(x_1,y_1)$ with $\rho(0) = (x_0, y_0)$ and $\rho(1)=(x_1,y_1)$. If you let $$D = \sqrt{(x_1-x_0)^2 + (y_1-y_0)^2},$$ then it is fairly easy to show that

$$|\rho(t) - \rho(s)| = D|t-s|$$

So, if you want to find the points a distance of $d$ from $(x_0,y_0)$, then you need to solve $$d = |\rho(t) - \rho(0)| = D|t-0|$$

You get $ t = \pm \dfrac dD$. Hence the points are

$$\rho\left( \pm \dfrac dD \right) = \left\{ \begin{array}{c} \left( 1-\dfrac dD \right)(x_0,y_0) + \dfrac dD (x_1,y_1) \\ \left( 1+\dfrac dD \right)(x_0,y_0) - \dfrac dD (x_1,y_1) \end{array} \right.$$

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