Chapter 4.6, Problem 5PSA

a.

To determine

To find: Slope of the given line.

Answer to Problem 5PSA

  $\frac{1}{4}$ .

Explanation of Solution

Given information:

A figure is given as below.

  

Calculation:

A figure is given as below.

  

  $\text{Slope of }\overleftrightarrow{BE}=\frac{△y}{△x}=\frac{3-1}{11-3}=\frac{1}{4}.$

b.

To determine

To find: Slope of the given line.

Answer to Problem 5PSA

  $-4$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

A figure is given as below.

  

  $\begin{array}{l}\text{Slope of }\overleftrightarrow{BE}=\frac{△y}{△x}=\frac{3-1}{11-3}=\frac{1}{4}.\\ \text{Since }\overleftrightarrow{AC}\text{ is perpendicular to }\overleftrightarrow{BE}.\\ \Rightarrow \text{Slope of }\overleftrightarrow{AC}=-\frac{1}{\text{Slope of }\overleftrightarrow{BE}}\\ \Rightarrow \text{Slope of }\overleftrightarrow{AC}=-\frac{1}{\frac{1}{4}}\\ \Rightarrow \text{Slope of }\overleftrightarrow{AC}=-4\end{array}$

c.

To determine

To find: Slope of the given line.

Answer to Problem 5PSA

  $-\frac{7}{2}$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

A figure is given as below.

  

Using midpoint formula,

  $D=\left(\frac{11+3}{2},\frac{3+1}{2}\right)=\left(7,2\right)$

  $\text{Slope of }\overleftrightarrow{AD}=\frac{△y}{△x}=\frac{2-9}{7-5}=-\frac{7}{2}$

d.

To determine

To find: Slope of the given line.

Answer to Problem 5PSA

  $\frac{1}{4}$ .

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

A figure is given as below.

  

Slope of a line passing through A and parallel to the line $\overleftrightarrow{BE}$ will have same slope as of slope of line $\overleftrightarrow{BE}$

  $\text{Slope of }\overleftrightarrow{BE}=\frac{△y}{△x}=\frac{3-1}{11-3}=\frac{1}{4}.$