Let $a,b\in\mathbb R^2$ be non-collinear and let $z\in\mathbb R^2$. After having proven that $ \exists \lambda,\mu\in\mathbb R \mid \lambda a+\mu b=z$ using simultaneous equations academically, I tried to find a more visual proof.
For a start, I know that it is necessary that $z-\lambda a$ and $b$ be collinear, and therefore we must have $\det(z-\lambda a,b)=0$. Solving for $\lambda$ using coordinates gives Cramer's formula : $\lambda = \frac{\det(z,b)}{\det(a,b)}$. I realise here that from symmetry arguments you can deduce that it is also necessary that $\mu=\frac{\det(z,a)}{\det(b,a)}$. However, if I'm not mistaken, so far these conditions are necessary and need to be proven sufficient by computing $\lambda a + \mu b$ using coordinates, which I'm convinced may be a waste of time.
Attempting to circumvent the use of coordinates once more, what we do know about $\lambda$ and $\mu$ is that $z-\lambda a$ and $\mu b$ are collinear, and so are the pair $z-\mu b$ and $\lambda a$. I have a feeling that this in itself is sufficient to prove that $\lambda a+\mu b=z$ concisely and in an elementary way, could anyone help with this ?
From the given co-linearities, we have
$$z-a=tb,\>\>\>\>\>z-b = sa \tag 1$$
Take the difference of the two equations to get $a-b= sa -tb$, or
$$(1-s)a=(1-t)b$$
Given that $a$ and $b$ are non-colinear, we have $t=s=1$ for above equation to hold. Plug into (1) and take the sum of the two equations to get
$$2z -(a+b) = a+b$$
which yields $z=a+b$.