Ratio Theorem

If \(P\) divides \(AB\) internally in the ratio \(AP:PB=\lambda:\mu\), where \(\lambda,\mu>0\), then

For external division, use directed ratios consistently; the denominator becomes a difference and must be non-zero.

Derive a division formula from movement along the segment: \(\mathbf p=\mathbf a+t(\mathbf b-\mathbf a)=(1-t)\mathbf a+t\mathbf b\). For \(AP:PB=m:n\), \(t=\dfrac{m}{m+n}\). The larger weight multiplies the endpoint closer to \(P\).

The parameter locates the point: \(0<t<1\) is inside \(AB\), \(t=0,1\) are its endpoints, and \(t<0\) or \(t>1\) is outside. For example, \(AP:PB=2:3\) gives \(\mathbf p=\dfrac{3\mathbf a+2\mathbf b}{5}\).

For an external point, first draw the order of the points and express its displacement as a signed multiple of \(\overrightarrow{AB}\). Check the resulting parameter against that order. This avoids swapping the weights or confusing \(AP:PB\) with \(AP:AB\).

Typed solution erratum

Correction to the accompanying typed example: when A divides PQ internally with \(PA:AQ=1:3\), \(\overrightarrow{OA}=\overrightarrow{OP}+\dfrac14(\overrightarrow{OQ}-\overrightarrow{OP})=\dfrac{3\overrightarrow{OP}+\overrightarrow{OQ}}4\). The typed example now uses these corrected weights. The larger weight belongs to P, the closer endpoint. The general ratio theorem above is correct; the original recorded example and media are retained.