In triangle $OAB$, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. Poin

Question

In triangle $OAB$, $\overrightarrow{OA}=\mathbf{a}$ and $\overrightarrow{OB}=\mathbf{b}$. Point $P$ has position vector $\overrightarrow{OP}=\dfrac13\mathbf{a}+\dfrac23\mathbf{b}$. Show that $P$ lies on the segment $AB$ and find $AP:PB$.[3] Find the ratio of the areas of triangles $OAP$ and $OPB$, explaining your answer.[2]

Accepted Answer

(a) $\overrightarrow{AP}=\dfrac23\overrightarrow{AB}$; $2:1$ (b) $2:1$, common height

A division ratio and triangle areas

A division ratio and triangle areas

Solution:

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