# Find the point that divides the line segment $3/5$ of the way

Given $A(-6,4)$ and $B(19,29)$, find the point that divides the line segment $AB$ $3/5$ of the way from $A$ to $B$.

Since the line is not horizontal or vertical, it is a little more difficult. I found the midpoint of the line, which is $(13,25)$ but I still need to go $0.5$ of the way more.

The distance of the line is $\sqrt{1250}$ but I am not sure how or if I am able to use this information.

Any help would be appreciated! Thank you!

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You found the midpoint by going $\frac{1}{2}$ of the way from $-6$ to $19$, and $\frac{1}{2}$ of the way from $4$ to $29$. Similarly, if you want the point $\frac{3}{5}$ of the way from $A$ to $B$, you want to go $\frac{3}{5}$ of the way from $-6$ to $19$ (which is $9$ [Why?]) and from $4$ to $29$ (which is $19$ [Why?]). So the point is $(9,19)$.

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Thank you! This makes sense! –  SamHaim Jun 12 '14 at 16:04
No problem. If the answer satisfies you, don't forget to accept it by checking the check below the up/down arrows. –  rogerl Jun 12 '14 at 16:04

The coordinate of a Point $P(x,y)$ that divides the two points $A(x_1,y_1), B(x_2,y_2)$ internally in the ratio $m:n$

then $$\frac{x-x_2}n=\frac{x_1-x}m\iff x=\frac{mx_2+nx_1}{m+n}$$

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This can be done with vector arithmetic, meaning scalar multiplication and vector addition.

Given two points $A,B$ in the coordinate plane (where we think of $A,B$ as the tips of vectors based at the origin), and given a number $t \in [0,1]$, the point which is "$t$ of the way from $A$ to $B$" is given by the following vector formula which combines scalar multiplication with vector addition: $$(1-t) A + t B$$ Intuitively, what you are doing is marking off the line segment $\overline{AB}$ by points of the unit interval $[0,1]$. The formula is just a way of translating this into vector arithmetic.

So the answer to your question is $$\biggl(1 - \frac{3}{5}\biggr) (-6,4) + \biggl(\frac{3}{5}\biggr) (19,29)$$ and the rest is vector arithmetic.

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