# Devide line to 3 points

I have two point in a coordinates system, let's say $(x_1,y_1)$ and $(x_2,y_2)$, and I want to find the coordinates of the point that separates the line into 3 parts

I want to know the coordinates of the two points, I tried this formula on them $Pt_x = (x_1+x_2)/3$ and $Pt_y = (y_1+y_2)$ but it didn't work

• you can parametrize the line that goes through these two both points and then plug in $1/3(x_2-x_1)+y_1$ and $2/3(x_2-x_1)+y_1$ for x – math635 Mar 16 '16 at 21:01
• I can't do this in my case – Marox Tn Mar 16 '16 at 21:06
• Is there any straightforward formula ? – Marox Tn Mar 16 '16 at 21:06
• You can see in IanF1 answer what I meant – math635 Mar 16 '16 at 21:28

Hint:

If $A=(x_1,y_1)$ and $B=(x_2,y_2)$ and you want a point $P=(x,y)$ (on the line $AB$) such that $\overline{AP}=k\overline{AB}$ than, by similitude of triangles, you can esily found that: $$x=x_1+k(x_2-x_1) \qquad y=y_1+k(y_2-y_1)$$

can you use this in your case?

• Yeah it's working fine, thanks ! – Marox Tn Mar 16 '16 at 21:18

You can start at the point $(x_1,y_1)$ and from there move $1/3$ of the total difference in the $x$-coordinates and $1/3$ of the total difference in the $y$-coordinates, so you end up in the point $(x_1 + 1/3(x_2-x_1) , y_1 + 1/3(y_2-y_1)$. Do that again and you obtain $(x_1 + 2/3(x_2-x_1) , y_1 + 2/3(y_2-y_1)$. Making a third step would result in $(x_1 + 3/3(x_2-x_1) , y_1 + 3/3(y_2-y_1) = (x_2,y_2)$, just like you would expect.

The line from $(x_1, y_1)$ to $(x_2, y_2)$ can always be parameterised as $((1-t)x_1 + tx_2, (1-t)y_1 + ty_2)$.

Plugging in $t=0$ gives back $(x_1, y_1)$ and plugging in $t=1$ gives $(x_2, y_2)$.

The coordinates of points any fraction of the way along the line can be found by setting t = that fraction. So in your case you need $t = \frac{1}{3}$ and $t=\frac{2}{3}$.

Let $A=(x_1,y_1),B(x_3,y_3),C(x_4,y_4),D(x_2,y_2)$

You know the line in which $A,D$ lie, let's call it $l:y=ax+b$ (you can find the parameters $a,b$) and the fact that both B,C are in that line as well. So, the coordinates of $B,C$ need to satisfy the equation of the line. $$y_3=ax_3+b\\y_4=ax_4+b$$

Also, you know the length of $|AD|=L$. You require $$|AB|=\sqrt{(x_1-x_3)^2+(y_1-y_3)^2)}=\\|CD|=\sqrt{(x_2-x_4)^2+(y_2-y_4)^2)}\\=\frac{L}{3}$$
So you have a system that will allow you to find $B,C$.

If you are seeking middle two coordinates of $P,Q$ by proportion

$x_P = (2 x_1+x_2)/3$ and

$y_P = (2 y_1+y_2)/3$

$x_Q = ( x_1+ 2x_2)/3$ and

$y_Q = ( y_1+2 y_2)/3.$