Chapter 5.4, Problem 3P

a.

To determine

To write: an equation of a line using point (-2, -3).

Answer to Problem 3P

  $y+3=\frac{7}{3}(x+2)$

Explanation of Solution

Given information: an equation in the last step of above example is given as follows-

  $y-4=\frac{7}{3}(x-1)$

We have to write an equation using point (-2, -3) instead of (1, 4).

  $\begin{array}{l}y-(-3)=\frac{7}{3}(x-(-2))\\ \Rightarrow y+3=\frac{7}{3}(x+2)\end{array}$

b.

To determine

To write: the equations in slope intercept form and compare them.

Answer to Problem 3P

  $y=\frac{7}{3}x+\frac{5}{3}$

Explanation of Solution

Given information: an equation in the last step of above example is given as follows-

  $y-4=\frac{7}{3}(x-1)$

An equation in the last step of above example is given as follows-

  $y-4=\frac{7}{3}(x-1)$

  $\begin{array}{l}\Rightarrow y-4=\frac{7}{3}x-\frac{7}{3}\\ \Rightarrow y=\frac{7}{3}x-\frac{7}{3}+4\\ \Rightarrow y=\frac{7}{3}x+\frac{5}{3}\end{array}$

This is the required equation in slope intercept form.

Now the equation obtained in part a is as follows-

  $\begin{array}{l}y+3=\frac{7}{3}(x+2)\\ \Rightarrow y+3=\frac{7}{3}x+\frac{14}{3}\\ \Rightarrow y=\frac{7}{3}x+\frac{14}{3}-3\\ \Rightarrow y=\frac{7}{3}x+\frac{5}{3}\end{array}$

Here the both equations are same.

The point (-2, -3) already lies on the line in last step above example.

We can conclude that we can use any point on a line to write an equation of the line in point slope form.