Given the points $A(2a, a)$ and $B(2b, b)$ find the coordinates of the point $M$ such that $\vec{AM} = 3 \vec{MB}$.

Question

Consider the points:

$$A(2a, a) \hspace{2cm} B(2b, b)$$

with $a \ne b$ and $a, b \in \mathbb{R}$. Find the point $M(x, y)$ such that $\vec{AM} = 3 \vec{MB}$.

First thing I tried was to plot everything we know so far. Here's what the graph looks like with $a = 2$ and $b = 4$:

So it's clear that what I have to do is find the point $M(x, y)$ on the segment $[AB]$ such that we have $\vec{AM} = 3 \vec{MB}$. I thought that an interactive graph would be helpful to see how thing move around, so here it is.

What I did first was create the vectors

$$\vec{OA} = \begin{pmatrix} 2a \\ a \end{pmatrix} \hspace{2cm} \vec{OB} = \begin{pmatrix} 2b \\ b \end{pmatrix}$$

Then I got the vector $\vec{AB}$:

$$\vec{AB} = \vec{OB} - \vec{OA} = \begin{pmatrix} 2(b - a) \\ b - a \end{pmatrix}$$

Now since we need $\vec{AM} = 3\vec{MB}$ that means that the point $\vec{AM}$ is situated at three quarters length of the whole vector $\vec{AB}$. So since we now have the vector $\vec{AB}$ we just need to shorten this vector, while keeping its direction. So I created the matrix:

$$X = \begin{pmatrix} \dfrac{3}{4} & 0 \\ 0 & \dfrac{3}{4} \end{pmatrix}$$

The matrix $X$ squishes each vector to three quarters of its length.

So now we have:

$$\vec{AM} = X \cdot \vec{AB} = \begin{pmatrix} \dfrac{3}{4} & 0 \\ 0 & \dfrac{3}{4} \end{pmatrix} \begin{pmatrix} 2(b - a) \\ b - a \end{pmatrix} = \begin{pmatrix} \dfrac{3(b - a)}{2} \\ \dfrac{3(b - a)}{4} \end{pmatrix}$$

So we can now find the coordinates of $M$.

$$\vec{OM} = \vec{OA} + \vec{AM} = \begin{pmatrix} 2a \\ a \end{pmatrix} + \begin{pmatrix} \dfrac{3(b - a)}{2} \\ \dfrac{3(b - a)}{4} \end{pmatrix} = \begin{pmatrix} \dfrac{a+3b}{2} \\ \dfrac{a+3b}{4} \end{pmatrix}$$

So we can conclude that the point $M$ is at:

$$M \bigg ( \dfrac{a+3b}{2}, \dfrac{a+3b}{4} \bigg )$$

My question is: Is this correct? We've never used matrices like this before in class, it's just an idea that came to me, so I have no idea if my work is correct, even though it seems right. And if it is correct, what is another way of solving this problem? I couldn't come up with any other solution, so that's why I had to do this "slight of hand".

Answer 1

It'd be simpler to operate in vector space directly. Note,

$$\vec{AB}=\vec{AM} + \vec{MB} = \vec{AM} + \frac13 \vec{AM} = \frac43 \vec{AM}\implies \vec{AM} = \frac34 \vec{AB}$$

Thus,

$$\vec{OM} = \vec{OA} + \vec{AM} =\vec{OA} + \frac34 \vec{AB} = \vec{OA} + \frac34 (\vec{OB} - \vec{OA}) = \frac14 \vec{OA} + \frac 34 \vec{OB} $$ $$=\frac14(2a, a) + \frac34 (2b,b)= \left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)$$

Answer 2

What you're doing looks correct. Using vectors and matrices is a valid way to proceed. However, as you requested, here is an alternate method to consider.

The line going through $A$ and $B$ has a slope of

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - a}{2b - 2a} = \frac{1}{2} \tag{1}\label{eq1A}$$

Thus, the general equation for it would be

$$y = \left(\frac{1}{2}\right)x + c \tag{2}\label{eq2A}$$

Using $y = a$ and $x = 2a$ gives that $c = 0$, so $y = \left(\frac{1}{2}\right)x \implies x = 2y$. Note this can be parametrized by having $y = a + t$ so $x = 2a + 2t$, giving

$$f(t) = (2a + 2t, a + t) \tag{3}\label{eq3A}$$

with $f(0) = A(2a, a)$ and $f(b - a) = B(2b, b)$. Since $M$ is three-quarters of the way between $A$ and $B$, then the point is

$$\begin{equation}\begin{aligned} M(x,y) & = f\left(\left(\frac{3}{4}\right)\left(b-a\right)\right) \\ & = \left(2a + 2\left(\frac{3}{4}\right)\left(b-a\right), a + \left(\frac{3}{4}\right)\left(b-a\right)\right) \\ & = \left(\frac{4a}{2} + \frac{3b-3a}{2}, \frac{4a}{4} + \frac{3b-3a}{4}\right) \\ & = \left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right) \end{aligned}\end{equation}\tag{4}\label{eq4A}$$

This matches your answer of $M\left(\frac{a + 3b}{2}, \frac{a + 3b}{4}\right)$.