Chapter 4.6, Problem 8PSA

a.

To determine

To find: If $\overleftrightarrow{RE}\parallel \overleftrightarrow{TC}.$

Answer to Problem 8PSA

  $\overleftrightarrow{RE}\parallel \overleftrightarrow{TC}$ .

Explanation of Solution

Given information:

A figure is given as below.

  

Concept used: Slope of a line passing through two points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given as $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$ .

Slope of two parallel lines are equal.

Calculation:

  

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{RE}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-1}{10-\left(-2\right)}=\frac{1}{3}\\ \text{Slope of line }\overleftrightarrow{TC}=\frac{y_2-y_1}{x_2-x_1}=\frac{8-4}{9-\left(-3\right)}=\frac{1}{3}\\ \text{Therefore, }\overleftrightarrow{RE}\parallel \overleftrightarrow{TC}.\end{array}$

In above figure,

b.

To determine

To find: If $\overleftrightarrow{TR}\parallel \overleftrightarrow{CE}.$

Answer to Problem 8PSA

  $\overleftrightarrow{TR}\parallel \overleftrightarrow{CE}.$

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Concept used: Slope of a line passing through two points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given as $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$ .

Slope of two parallel lines are equal.

Calculation:

  

In above figure,

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{TR}=\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{-2-\left(-3\right)}=-3\\ \text{Slope of line }\overleftrightarrow{CE}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-8}{10-9}=-3\\ \text{Therefore, }\overleftrightarrow{TR}\parallel \overleftrightarrow{CE}.\end{array}$

c.

To determine

To find: If $\angle R$ is a right angle.

Answer to Problem 8PSA

  $\angle R$ is a right angle.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Concept used: Slope of a line passing through two points $\left(x_1,y_1\right)\text{ and }\left(x_2,y_2\right)$ is given as $m=\frac{y_2-y_1}{x_2-x_1}=\frac{y_1-y_2}{x_1-x_2}$ .

Slope of two parallel lines are equal.

Calculation:

  

In above figure,

  $\begin{array}{l}\text{Slope of line }\overleftrightarrow{TR}=\frac{y_2-y_1}{x_2-x_1}=\frac{1-4}{-2-\left(-3\right)}=-3\\ \text{Slope of line }\overleftrightarrow{RE}=\frac{y_2-y_1}{x_2-x_1}=\frac{5-1}{10-\left(-2\right)}=\frac{1}{3}\\ \text{Since, Slope of line }\overleftrightarrow{TR}=-\frac{1}{\text{Slope of line }\overleftrightarrow{RE}}\\ \text{Therefore, }\overleftrightarrow{TR}\perp \overleftrightarrow{CE}.\end{array}$

Hence, $\angle R$ is a right angle.