A is the point (2,3) and B is the point (3,-5).

1. Find the gradient of AB.

For $A(2,3)$ and $B(3,-5)$,
$m=\frac{-5-3}{3-2}=\frac{-8}{1}=-8.$

2. Use $y=mx+c$ and one point to find the equation.

Using $y=mx+c$ with $m=-8$ and point $(2,3)$,
$3=-8(2)+c$
$3=-16+c$
$c=19.$
Therefore,
$y=-8x+19.$

1. Substitute the coordinates into the distance formula.

For $A(2,3)$ and $B(3,-5)$,

$AB=\sqrt{(3-2)^2+(-5-3)^2}$

2. Evaluate the length exactly.

$AB=\sqrt{1^2+(-8)^2}$

$AB=\sqrt{1+64}=\sqrt{65}$

1. Interpret the vector from B to C as a displacement.

$B=(3,-5)$ and
$\overrightarrow{BC}=\begin{pmatrix}-4\\3\end{pmatrix}.$
So from $B$ to $C$, move $-4$ in the $x$-direction and $+3$ in the $y$-direction.

2. Add the displacement to the coordinates of B to find C.

$C=(3+(-4),\,-5+3)$
$C=(-1,-2).$

1. Express the diagonal vector of the parallelogram as the sum of the two adjacent side vectors.

In parallelogram $OABC$,
$\overrightarrow{OB}=\overrightarrow{OA}+\overrightarrow{OC}.$
Given $\overrightarrow{OA}=2\mathbf{a}$ and $\overrightarrow{OC}=3\mathbf{c}$,
$\overrightarrow{OB}=2\mathbf{a}+3\mathbf{c}.$

1. Use the ratio $OT:TB=2:1$ to find the fraction of $\overrightarrow{OB}$ represented by $\overrightarrow{OT}$.

Since $OT:TB=2:1$, point $T$ is $\dfrac{2}{3}$ of the way from $O$ to $B$.

Using $\overrightarrow{OB}=2\mathbf{a}+3\mathbf{c}$,
$\overrightarrow{OT}=\frac{2}{3}\overrightarrow{OB}=\frac{2}{3}(2\mathbf{a}+3\mathbf{c})=\frac{4}{3}\mathbf{a}+2\mathbf{c}.$

1. Find the position vector of M using the midpoint of BC.

Since $M$ is the midpoint of $BC$,

$\overrightarrow{OM}=\frac{1}{2}\left(\overrightarrow{OB}+\overrightarrow{OC}\right)$

Using $\overrightarrow{OB}=2\mathbf a+3\mathbf c$ and $\overrightarrow{OC}=3\mathbf c$,

$\overrightarrow{OM}=\frac{1}{2}(2\mathbf a+3\mathbf c+3\mathbf c)$

$\overrightarrow{OM}=\mathbf a+3\mathbf c$

2. Express vectors from A to T and from A to M.

Given $\overrightarrow{OA}=2\mathbf a$ and from the earlier part,

$\overrightarrow{OT}=\frac{4}{3}\mathbf a+2\mathbf c$

Then

$\overrightarrow{AT}=\overrightarrow{OT}-\overrightarrow{OA}$

$=\left(\frac{4}{3}\mathbf a+2\mathbf c\right)-2\mathbf a$

$=-\frac{2}{3}\mathbf a+2\mathbf c$

Also,

$\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}$

$=(\mathbf a+3\mathbf c)-2\mathbf a$

$=-\mathbf a+3\mathbf c$

3. Check whether the vectors are parallel and conclude collinearity.

Compare $\overrightarrow{AT}$ and $\overrightarrow{AM}$:

$\frac{2}{3}\overrightarrow{AM}=\frac{2}{3}(-\mathbf a+3\mathbf c)$

$=-\frac{2}{3}\mathbf a+2\mathbf c$

$=\overrightarrow{AT}$

Hence $\overrightarrow{AT}$ is a scalar multiple of $\overrightarrow{AM}$. Therefore, A, T and M lie on a straight line.