I'm going through Multivariable and Vector Calculus by Sarhan Musa and David Santos, and I'm stuck on exercise 1.1.9.
The question says:
Let A,B be two points on the plane. Construct two points I and J such that $$\overline{IA}=-3\overline{IB}$$ $$\overline{JA} = -\frac{1}{3}\overline{JB}$$ and then demonstrate that for any arbitrary point M on the plane $$\overline{MA} + 3\overline{MB} = 4\overline{MI}$$ and $$3\overline{MA} + \overline{MB} = 4\overline{MJ}$$
I've drawn the question out and I can see vaguely geometrically how that could be the case. But I don't see how the author arrives at the answer (from the back of the book):
$$\begin{align}\overline{MA} + 3\overline{MB} &= 3\overline{MI}+\overline{MI}+\overline{IA}+3\overline{IB} \tag{1}\\&= 4\overline{MI} + \overline{IA} + 3\overline{IB}\tag{2}\\&= 4\overline{MI}\tag{3} \end{align}$$ and $$\begin{align}3\overline{MA} + \overline{MB} &= 3\overline{MJ} + 3\overline{JA}+\overline{MJ}+\overline{JB} \tag{4}\\ \tag{5}&=4\overline{MJ} + 3\overline{JA} + \overline{JB} \\ \tag{6}&= 4\overline{MJ}\end{align}$$
Specifically, how does the author arrive at the right hand side of lines 1 and 4?
For (1), we write: \begin{align*}\overline{MA} + 3\overline{MB} &= \left(\overline{MI}+\overline{IA} \right) + 3\left(\overline{MI}+\overline{IB}\right)\\ &= \overline{MI}+\overline{IA} + 3\overline{MI} + 3\overline{IB}\\ &= 3\overline{MI} + \overline{MI}+\overline{IA} + 3\overline{IB} \end{align*} The same computation with $J$ instead of $I$ yields (4)