Chapter 4.6, Problem 9PSA

a.

To determine

To find: The slope of line $\overleftrightarrow{PT}$ .

Answer to Problem 9PSA

  $\frac{3}{2}$ .

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:

  

In above figure,

  $\text{Slope of line }\overleftrightarrow{PT}=\frac{y_2-y_1}{x_2-x_1}=\frac{6-3}{4-2}=\frac{3}{2}$

Hence, slope of line $\overleftrightarrow{PT}$ is $\frac{3}{2}$ .

b.

To determine

To find: The slope of line $\overleftrightarrow{TV}$ .

Answer to Problem 9PSA

  $\frac{1}{2}$ .

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,

  $\text{Slope of line }\overleftrightarrow{TV}=\frac{y_2-y_1}{x_2-x_1}=\frac{8-6}{8-4}=\frac{1}{2}$

Hence, slope of line $\overleftrightarrow{TV}$ is $\frac{1}{2}$ .

c.

To determine

To find: If point P, T and V are collinear.

Answer to Problem 9PSA

Point P, T and V are non-collinear.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

  $\text{Slope of line }\overleftrightarrow{TV}=\frac{1}{2}$

  $\text{Slope of line }\overleftrightarrow{PT}=\frac{3}{2}$

  $\text{Since, Slope of line }\overleftrightarrow{PT}\ne \text{Slope of line }\overleftrightarrow{TV}$ .

Therefore, points P, T and V are non-collinear.