Chapter 13.4, Problem 30WE

a.

To determine

To calculate: Vector $\overrightarrow{AB}$ and $\overrightarrow{AM}$ and $\overrightarrow{AT}$ .

Answer to Problem 30WE

Vector $\overrightarrow{AB}=\left(30,-24\right)$ , $\overrightarrow{AM}=\left(15,-21\right)$ , $\overrightarrow{AT}=\left(10,-8\right)$ .

Explanation of Solution

Given information: In the diagram, M is the mid-point of $\overrightarrow{AB}$ and T is a trisector point of $\overrightarrow{AB}$ .

  

In the diagram, M is the mid-point of $\overrightarrow{AB}$

Vector

  $\begin{array}{l}\overrightarrow{AB}=\left(20+10,-15-9\right)\\ \overrightarrow{AB}=\left(30,-24\right)\end{array}$

M is the mid-point of $\overrightarrow{AB}$

Then

  $\begin{array}{l}M=\left(\frac{20-10}{2},\frac{-15-9}{2}\right)\\ M=\left(5,-12\right)\end{array}$

Vector $\overrightarrow{AM}$

  $\begin{array}{l}\overrightarrow{AM}=\left(5+10,-12-9\right)\\ \overrightarrow{AM}=\left(15,-21\right)\end{array}$

Trisector means dividing a line segment in the ratio $1:2$ and $2:1$

  $\begin{array}{l}T=\left(\frac{1\times 20-2\times 10}{3},\frac{-1\times 15+2\times 9}{3}\right)\\ T=\left(0,1\right)\end{array}$

Vector $\overrightarrow{AT}$

  $\overrightarrow{AT}=\left(10,-8\right)$

b.

To determine

To calculate: The co-ordinates of M and T .

Answer to Problem 30WE

M is the mid-point of $\overrightarrow{AB}$

Then

  $M=\left(5,-12\right)$

The co-ordinates of $T=\left(0,1\right)$ .

Explanation of Solution

Given information: In the diagram, M is the mid-point of $\overrightarrow{AB}$ and T is a trisector point of $\overrightarrow{AB}$ .

  

In the diagram, M is the mid-point of $\overrightarrow{AB}$

Vector

  $\begin{array}{l}\overrightarrow{AB}=\left(20+10,-15-9\right)\\ \overrightarrow{AB}=\left(30,-24\right)\end{array}$

M is the mid-point of $\overrightarrow{AB}$

Then

  $\begin{array}{l}M=\left(\frac{20-10}{2},\frac{-15-9}{2}\right)\\ M=\left(5,-12\right)\end{array}$

Trisector means dividing a line segment in the ratio $1:2$ and $2:1$

  $\begin{array}{l}T=\left(\frac{1\times 20-2\times 10}{3},\frac{-1\times 15+2\times 9}{3}\right)\\ T=\left(0,1\right)\end{array}$