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I am working on my scholarship exam practice (high school/pre-university level) and stuck at question (2) below.

Let $A, B, C$ be three points on a plane and $O$ be the origin point on this plane. Put $\vec{a}=\vec{OA}$, $\vec{b}=\vec{OB}$, $\vec{c}=\vec{OC}$. $P$ is a point inside the triangle $ABC$. Suppose that the ratio of the areas of $\Delta PAB$, $\Delta PBC$, $\Delta PCA$ is $2:3:5$.

(1) The straight line $BP$ intersects the side $AC$ at the point $Q$. Find $AQ:QC$.

My answer is $2:3$ and it is in accordance with the answer key. The ratio was obtained from the area ratio of $\Delta PAB$ and $\Delta PBC$ which share the same base.

$Area=\frac{1}{2}(base)(height)$

So the height of both $AQ$ and $QC$ is directly proportional to its area.

My problematic question is:

(2) Express $\vec{OP}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{c}$.

I honestly do not know how I should begin here. The answer key to this question is $\frac{1}{10}(3\vec{a}+5\vec{b}+2\vec{c})$. Please help.

Note: below is the actual answer key but it is in Japanese, just for reference. I do not know how we can explain from it though.

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  • $\begingroup$ See en.wikipedia.org/wiki/Barycentric_coordinate_system $\endgroup$ – amd Jun 18 at 18:49
  • $\begingroup$ Thank you for your link. I am still struggling at how we can choose the coefficients for $\vec{BA}$ and $\vec{BC}$ as $\frac{3}{5}$ and $\frac{2}{5}$, respectively. I am aware that all coefficients must add up to $1$ but not sure where those combinations came from. I am trying to find other websites to explain about this but most examples just show the numbers out of nowhere without further explanation. $\endgroup$ – Trey Anupong Jun 19 at 9:25
  • $\begingroup$ Hi thank you again for your suggestion. I found a nice YouTube video which is suitable for beginners, IMO. So problem solved now :) youtube.com/watch?v=G_8e5f8iNBw $\endgroup$ – Trey Anupong Jun 19 at 13:18

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