Chapter 13.4, Problem 4CE

a.

To determine

To explain, is $\overrightarrow{AG}$ parallel to $\overrightarrow{OB}$ .

Answer to Problem 4CE

From calculation part we get, $m_{AG}=m_{OB}$ parallel.

Explanation of Solution

Given information: The graph is.

  

Calculation: The required solution is obtained as,

slope $AG=\frac{y_2-y_1}{x_2-x_1}=\frac{0-\left(-3\right)}{-3-1}=-\frac{3}{4}$

slope $OB=\frac{y_2-y_1}{x_2-x_1}=\frac{-3-0}{4-0}=-\frac{3}{4}$

Hence, from above $m_{AG}=m_{OB}$ parallel.

b.

To determine

To explain, Is $\overrightarrow{AG}=\overrightarrow{OB}$ .

Answer to Problem 4CE

From calculation part we get, $\overrightarrow{AG}=\overrightarrow{OB}$ .

Explanation of Solution

Given information: The graph is.

  

Calculation: The required solution is obtained as,

  $\begin{array}{l}\overrightarrow{AG}=\left(x_2-x_1,y_2-y_1\right)=\left(-3-1,0-\left(-3\right)\right)=\left(-4,3\right)\\ \overrightarrow{OB}=\left(x_2-x_1,y_2-y_1\right)=\left(4-0,-3-0\right)=\left(4,-3\right)\\ \overline{AG}=\overrightarrow{OB}=\sqrt{x^2+y^2}=\sqrt{4^2+3^2}=\sqrt{25}=5\\ \overrightarrow{AG}=\overrightarrow{OB}\end{array}$

Hence, from the above solution $\overrightarrow{AG}=\overrightarrow{OB}$ .