Chapter 4.6, Problem 14PSB

a.

To determine

To find: If the point J lie on $\overleftrightarrow{DE}$ .

Answer to Problem 14PSB

Yes.

Explanation of Solution

Given information:

A figure is given as below. Point $H\left(0,7\right)$ is rotated by ${90}^0$ about origin in clockwise direction.

  

Calculation:

A figure is given as below.

  

Point $H\left(0,7\right)$ is rotated by ${90}^0$ about origin in clockwise direction.

So, coordinates of point J would be $\left(7,0\right)$ .

  $\begin{array}{l}\text{Slope of }\overleftrightarrow{DE}=\frac{2-\left(-3\right)}{11-1}=\frac{1}{2}\\ \text{Slope of }\overleftrightarrow{JE}=\frac{2-0}{11-7}=\frac{1}{2}\\ \text{Since slope of both lines are equal and one point is also common}\text{.}\\ \text{This represents same line}\text{. Therefore, point J lie on }\overleftrightarrow{DE}.\end{array}$

b.

To determine

To find: If $\overline{FJ}\text{ is the altitude of }\overline{DE}$ .

Answer to Problem 14PSB

No.

Explanation of Solution

Given information:

Same as part $a$ of this exercise.

Calculation:

A figure is given as below.

  

Point $H\left(0,7\right)$ is rotated by ${90}^0$ about origin in clockwise direction.

So, coordinates of point J would be $\left(7,0\right)$ .

  $\begin{array}{l}\text{Slope of }\overleftrightarrow{DE}=\frac{2-\left(-3\right)}{11-1}=\frac{1}{2}\\ \text{Slope of }\overleftrightarrow{FJ}=\frac{6-0}{5-7}=-3\\ \text{Since slope of }\overleftrightarrow{FJ}\text{ is not the negative reciprocal of slope of }\overleftrightarrow{DE}.\\ \text{This means both lines are not perpendicular}\text{.Therefore, }\overline{FJ}\text{ is not the altitude of }\overline{DE}.\end{array}$