# Point dividing Line Segment between Two Points in Given Ratio

## Theorem

Let $A$ and $B$ be points whose position vectors relative to an origin $O$ of a Euclidean space are $\mathbf a$ and $\mathbf b$.

Let $\mathbf r$ be the position vector of a point $R$ on $AB$ which divides $AB$ in the ratio $m : n$.

Then:

$\mathbf r = \dfrac {n \mathbf a + m \mathbf b} {m + n}$

## Proof 1

We have that:

$\vec {A B} = \mathbf b - \mathbf a$

and so:

$\vec {A R} = \dfrac m {m + n} \paren {\mathbf b - \mathbf a}$

Hence the position vector $\mathbf r$ of $R$ is given by:

 $\ds \mathbf r$ $=$ $\ds \vec {O R}$ $\ds$ $=$ $\ds \vec {O A} + \vec {A R}$ $\ds$ $=$ $\ds \mathbf a + \dfrac m {m + n} \paren {\mathbf b - \mathbf a}$ $\ds$ $=$ $\ds \dfrac {n \mathbf a + m \mathbf b} {m + n}$

$\blacksquare$

## Proof 2

Let the coordinates of $A$ be $\tuple {x_1, y_1}$.

Let the coordinates of $B$ be $\tuple {x_2, y_2}$.

Let the coordinates of $R$ be $\tuple {X, Y}$.

Then we have:

$\dfrac {x_2 - X} {X - x_1} = \dfrac n m$

and so:

$X \paren {m + n} = m x_2 + n x_1$

Similarly for $Y$, giving $R$ as:

$\tuple {\dfrac {m x_2 + n x_1} {m + n}, \dfrac {m y_2 + n y_1} {m + n} }$

The result follows.

$\blacksquare$