Chapter 4.2, Problem 54PFA

a)

To determine

To find: equations for line joining $\overline{EG}$ in slope point form.

Answer to Problem 54PFA

The point slope form of $\overline{EG}$ is $y-6=-\frac{5}{3}(x-3)$ .

Explanation of Solution

Given information:

Consider

From the graph the points are $E(-3,6),G(3,1)$

The line $\overline{EG}$ equation is

  $\begin{array}{l}\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\ \frac{y-6}{x-3}=\frac{1-6}{3+3}\\ \frac{y-6}{x-3}=-\frac{5}{3}\\ y-6=-\frac{5}{3}(x-3)\text{ point slope form}\end{array}$

Hence,

The point slope form of $\overline{EG}$ is $y-6=-\frac{5}{3}(x-3)$ .

b)

To determine

To write: the standard form of the line containing $\overline{EG}$ .

Explanation of Solution

Given information:

Consider

From the graph the points are $E(-3,6),G(3,1)$

The line $\overline{EG}$ equation is

  $\begin{array}{l}\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}\\ \frac{y-6}{x-3}=\frac{1-6}{3+3}\\ \frac{y-6}{x-3}=-\frac{5}{3}\\ 3y-18=5x-15\\ 5x-3y=3\text{ standard form }ax+by=c\end{array}$

Hence,

The standard form of $\overline{EG}$ is $5x-3y=3$ .