How came the coefficients of vectors in the Solution?

Question

Let (A,a1,a2) and (B,b1,b2) be two affine coordinate systems (illustrated below).Represent the point P in both systems. As Solution is in the Image . I want to Know that how are these values calculated e.g 2 ,3 ,5/2 and 3/2. Image

[1]: https://i.stack.imgur.com/SI3EE.png

Answer 1

Let be $O$ the center of a rectangular coordinate system fixed at the bottom-left corner of the grid as shown in the image:

The components of vectors $\vec{OA},\ \vec{OB}, \ \vec{OP}, \ \vec{a_1}, \ \vec{a_2}, \ \vec{b_1}$ and $\vec{b_2}$: $$ \vec{OA}=(2,1) $$

$$ \vec{OB}=(10,2) $$

$$ \vec{OP}=(6,3) $$

$$ \vec{a_1}=(-1,1) $$

$$ \vec{a_2}=(2,0) $$

$$ \vec{b_1}=\vec{a_1} $$

$$ \vec{b_2}=(-1,-1) $$

The goal is to express vector $\vec{OP}$ as the following linear combinations: $$ \vec{OP}=\vec{OA}+k_{1}\vec{a_1}+k_{2}\vec{a_2} \quad \textbf{[1]} $$ $$ \vec{OP}=\vec{OB}+m_{1}\vec{b_1}+m_{2}\vec{b_2} \quad \textbf{[2]} $$

From $\textbf{[1]}$, you get this linear system of equations: $$ 6=2-k_{1}+2k_{2} $$

$$ 3=1+k_{1} $$

Whose solution is $k_{1}=2$ and $k_{2}=3$

From $\textbf{[2]}$: $$ 6=10-m_{1}-m_{2} $$ $$ 3=2+m_{1}-m_{2} $$ You obtain $m_{1}=\frac{5}{2}$ and $m_{2}=\frac{3}{2}$