Divide a segment in a stated ratio

Divide a segment in a stated ratio

Sketch the point order and mark the two segment lengths first. For an internal point, the endpoint weights are positive and sum to one. For a point on a produced line, identify the middle point before applying the internal formula.

For distinct points $A,B$, write $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$. If $P$ divides $AB$ internally with $AP:PB=\lambda:\mu$, where $\lambda,\mu>0$, then the ratio theorem (listed in the source’s MF27) gives

$$\mathbf p=\overrightarrow{OP}=\frac{\mu\mathbf a+\lambda\mathbf b}{\lambda+\mu}.$$

The coefficient of each endpoint is the opposite segment’s share.

Derive the endpoint weights

Begin with the directed fraction of $AB$. The full derivation is:

$$\begin{aligned} \overrightarrow{AP}&=\frac\lambda{\lambda+\mu}\overrightarrow{AB},\\ \overrightarrow{AO}+\overrightarrow{OP}&=\frac\lambda{\lambda+\mu}(\overrightarrow{AO}+\overrightarrow{OB}),\\ -\mathbf a+\mathbf p&=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b),\\ \mathbf p&=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b)+\mathbf a\\ &=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b)+\frac{\lambda+\mu}{\lambda+\mu}\mathbf a\\ &=\frac{-\lambda\mathbf a+\lambda\mathbf b+\lambda\mathbf a+\mu\mathbf a}{\lambda+\mu}\\ &=\frac{\mu\mathbf a+\lambda\mathbf b}{\lambda+\mu}. \end{aligned}$$

All position vectors must start at the same point. That point need not be the origin: replacing $O$ by any common reference $C$ gives

$$\overrightarrow{CP}=\frac{\mu\overrightarrow{CA}+\lambda\overrightarrow{CB}}{\lambda+\mu}.$$

When $\lambda=\mu$, $P$ is the midpoint and $\mathbf p=(\mathbf a+\mathbf b)/2$. The source assumes nonzero endpoint position vectors, but the affine formula also works when an endpoint is the origin. The positive-ratio and distinct-endpoint conditions describe an internal division of a genuine segment.

Read “produced” in order

$AB$ produced continues from $A$ through $B$ and beyond $B$; $BA$ produced continues from $B$ through $A$ and beyond $A$. First sketch the order, then apply the internal theorem to the middle point. Its three cases are $A,P,B$ with $AP:PB=3:7$; $B,A,R$ with $AR:BR=2:7$ (so $BA:AR=5:2$); and $A,B,Q$ with $AQ:BQ=7:3$ (so $AB:BQ=4:3$). Example 6 contrasts an internal $2:3$ point with $AB$ produced at $AP:PB=2:1$, where $B$ is the midpoint of $AP$.