A point dividing a segment If X lies on AB with AX:XB=m:n, cut AB into m+n equal parts; AX is m of them. So AX= mm+n AB. The fraction is out of the whole segment AB, not out of XB. In the figure AX:XB=3:2, so AX= 35 AB. Method - Turn the ratio into a fraction of the whole: AX:XB=m:n gives AX= mm+n AB. - Check the direction: AX points the same way as AB, while XA=- mm+n AB. - Find AB= b- a (end minus start). - Add to the starting point: OX= OA+ AX. Numbers: a= 2 -1, b= 8 5 and AX:XB=1:2. Then AB= 6 6, AX= 13 6 6 = 2 2 and OX= 2 -1 + 2 2 = 4 1. Letters: with AX:XB=3:7, OX= a+ 310(b- a)= 710 a+ 310 b. Midpoints, reverse questions and points beyond the segment For the midpoint M of AB, m:n=1:1, so OM= a+ 12(b- a)= 12(a+ b). To find a ratio, write AX as a multiple of AB. If AX= 58 AB, then X is between A and B and AX:XB=5:3. If AP= 53 AB, the multiple is greater than 1, so P lies on AB produced beyond B, with BP= 23AB. Check A ratio of lengths becomes a vector equation only after the direction is fixed, and a vector is never divided by a vector: compare lengths, such as AX and AB, instead.

A point dividing a segment in a given ratio

Secondary 4

A point dividing a segment

If $X$ lies on $AB$ with $AX:XB=m:n$, cut $AB$ into $m+n$ equal parts; $AX$ is $m$ of them. So $\overrightarrow{AX}=\dfrac{m}{m+n}\overrightarrow{AB}.$

The fraction is out of the whole segment $AB$, not out of $XB$.

In the figure $AX:XB=3:2$, so $\overrightarrow{AX}=\dfrac35\overrightarrow{AB}$.

Method

Turn the ratio into a fraction of the whole: $AX:XB=m:n$ gives $\overrightarrow{AX}=\dfrac{m}{m+n}\overrightarrow{AB}$.
Check the direction: $\overrightarrow{AX}$ points the same way as $\overrightarrow{AB}$, while $\overrightarrow{XA}=-\dfrac{m}{m+n}\overrightarrow{AB}$.
Find $\overrightarrow{AB}=\mathbf{b}-\mathbf{a}$ (end minus start).
Add to the starting point: $\overrightarrow{OX}=\overrightarrow{OA}+\overrightarrow{AX}$.

Numbers: $\mathbf{a}=\begin{pmatrix}2\\ -1\end{pmatrix}$, $\mathbf{b}=\begin{pmatrix}8\\ 5\end{pmatrix}$ and $AX:XB=1:2$. Then $\overrightarrow{AB}=\begin{pmatrix}6\\ 6\end{pmatrix}$, $\overrightarrow{AX}=\dfrac13\begin{pmatrix}6\\ 6\end{pmatrix}=\begin{pmatrix}2\\ 2\end{pmatrix}$ and $\overrightarrow{OX}=\begin{pmatrix}2\\ -1\end{pmatrix}+\begin{pmatrix}2\\ 2\end{pmatrix}=\begin{pmatrix}4\\ 1\end{pmatrix}$.

Letters: with $AX:XB=3:7$, $\overrightarrow{OX}=\mathbf{a}+\dfrac3{10}(\mathbf{b}-\mathbf{a})=\dfrac7{10}\mathbf{a}+\dfrac3{10}\mathbf{b}$.

Midpoints, reverse questions and points beyond the segment

For the midpoint $M$ of $AB$, $m:n=1:1$, so $\overrightarrow{OM}=\mathbf{a}+\dfrac12(\mathbf{b}-\mathbf{a})=\dfrac12(\mathbf{a}+\mathbf{b})$.

To find a ratio, write $\overrightarrow{AX}$ as a multiple of $\overrightarrow{AB}$. If $\overrightarrow{AX}=\dfrac58\overrightarrow{AB}$, then $X$ is between $A$ and $B$ and $AX:XB=5:3$.

If $\overrightarrow{AP}=\dfrac53\overrightarrow{AB}$, the multiple is greater than 1, so $P$ lies on $AB$ produced beyond $B$, with $BP=\dfrac23AB$.

Check

A ratio of lengths becomes a vector equation only after the direction is fixed, and a vector is never divided by a vector: compare lengths, such as $AX$ and $AB$, instead.

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