Chapter 6.5, Problem 41PPE

To determine

Todetermine:The slope of boundary line of in inequality of the form $y>mx+b$ such that point $\left(1,2\right)$ is not a solution but $\left(3,2\right)$ is.

Answer to Problem 41PPE

The slope of boundary line of in inequality of the form $y>mx+b$ such that point $\left(1,2\right)$ is not a solution but $\left(3,2\right)$ is negative.

Explanation of Solution

Given information:

The point $\left(3,2\right)$ is the solution inequality of the form $y>mx+b$ but $\left(1,2\right)$ . The slope may be positive, negative, or undefined.

Methodused:

The slope of a line is defined as the tangent of the angle made by the line with x -axis. To determine the slope the graphical method is used.

The points $\left(3,2\right)$ and $\left(1,2\right)$ are graphed in Figure 1(a) and Figure-1(b). The boundary line of the inequality which has solution $\left(3,2\right)$ and $\left(1,2\right)$ not passes through between these points and the graph (shade) of the inequality include point $\left(3,2\right)$ but $\left(1,2\right)$ not.

  

There may be four lines passing between points $\left(3,2\right)$ and $\left(1,2\right)$ that of the form $y>mx+b$$y>-mx+b$ , $x>k$ and $x<k$ . For example graphs of $y>x+0.1$ and $y>-x+4$ shown in Figure 1(b) that shows inequality $y>-x+4$ whose slope is negative, includes the point $\left(3,2\right)$ but not to point $\left(1,2\right)$ .

Thus, the slope of boundary line of in inequality of the form $y>mx+b$ such that point $\left(1,2\right)$ is not a solution but $\left(3,2\right)$ is negative.

Conclusion:

The slope of boundary line of in inequality of the form $y>mx+b$ such that point $\left(1,2\right)$ is not a solution but $\left(3,2\right)$ is negative.