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Say I have a points $a=(1,1), b=(5,5), c=(4,4)$.

I know that point $c$ is $3/4$ of the way along segment $\overline{ab}$, but how would I work this out?

I know that the section formula can tell me where point $c$ is if I already have the ratio by which it divides line segment $\overline{ab}$, but I need to go the other way around.

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up vote 2 down vote accepted

Given a line segment with endpoints $a = (x_a, y_a), b = (x_b, y_b),\;$ and a point $c = (x_c, y_c\;$ lying on the line segment between end points $a$ and $b$ (assuming we're talking about points in the Cartesian plane):

You'll need to compute the distance $d_{a\to b}$ between points $a, b$, using the formula for Euclidean distance in $\mathbb R^2$): $$d_{a \to b}\sqrt{(x_a - x_b)^2 + (y_a - y_b)^2}$$

You'll then need to compute, in the same manner, the distance $d_{a\to c}$ between points $a$ and $ c,\;$ but with $x_c, y_c$ replacing the coordinates for point $d$ above to get $\;d_{a \to c}$

Then you need to divide: $\dfrac{d_{a \to c}}{d_{a \to b}}$. That will give you the ratio you need ratio.

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This is perfect, just what I'm looking for. It seems so simple now! Thanks for the help! – Fraserr Mar 12 '13 at 0:59
You're welcome! – amWhy Mar 12 '13 at 1:00

Consider the two points $P=(a,b)$ and $Q=(c,d)$. The points $(x,y)$ on the line segment joining $P$ and $Q$ have parametric representation $$x= a(1-t)+ct,\qquad y=(1-t)b+td,$$ where $t$ ranges from $0$ to $1$. To find the point which is $\frac{3}{4}$ of the way from $P$ to $Q$, use $t=\frac{3}{4}$.

If the points are already given to you, and you want to find the ratio, you need $t$. This can be found from either of the two equations above. For example, if we know the coordinates $(u,v)$ of our "partway" point, we set $u=a(1-t)+ct$, and find that $t=\frac{u-a}{c-a}$.

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Instead of points "$a,b,c$" let's call them $P_1=(x_1,y_1),P_2=(x_2,y_2),P_3=(x_3,y_3)$ then for the ratios we have (suppose $x_3$ is between the $x_1,x_2$) : $$x_3=\dfrac {w_1x_1+w_2x_2}{w_1+w_2}$$ solving for $w_1,w_2$ we get $$\dfrac {w_1}{w2}=\dfrac{x_2-x_3}{x_3-x_1}$$ in the case $x_1=1,x_2=5,x_3=4$

we get: $\dfrac {w_1}{w2}=\dfrac{x_2-x_3}{x_3-x_1}=\dfrac{5-4}{4-1}=\dfrac{1}{3}$

One can repeat the above process for $y_k$'s and get the same answer for $\dfrac {w_1}{w_2}$ or not, depending if the 3rd point is on the same line or not.

PS : Above I assumed $P_3$ is between $P_1,P_2$, but that was only to help with visualisation, there is no need for $P_3$ to be between $P_1,P_2$. Consider the case where $P_3$ is not on the same line, but the formula for the ratios is still valid.

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This avoids the need for square roots that arises if you compute the distances. – Ross Millikan Mar 12 '13 at 21:40
@RossMillikan : There is a rational geometry course on you tube that pretty much does everything without needing to use distances and irrationalities caused by them. – Arjang Mar 12 '13 at 21:49

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