# Given area of three points on a plane and $P$ is a point inside a triangle, find $\vec{OP}$

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.

• – amd Jun 18 at 18:49
• 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. – Trey Anupong Jun 19 at 9:25
• 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 – Trey Anupong Jun 19 at 13:18