Chapter 3.1, Problem 54PPSE

Solution

To determine the equation of the line that might be perpendicular to the line given in the graph from the options: $y=-3x+2,y=-\frac{1}{3}x-4,y=\frac{1}{3}x+5,\text{and }y=3x-1$ .

The line $y=\frac{1}{3}x+5$ seems to be the one that might be perpendicular to the line given in the graph.

Given:

Given the graph of a line:

  

Concept Used:

Slope of the line $m$passing through the points $\left(x_1,y_1\right)$and $\left(x_2,y_2\right)$:

  $m=\frac{y_2-y_1}{x_2-x_1}$ .

Slope intercept form of a line:

Given a line having the slope $m$ and $y$ -intercept $c$ , it has the equation $y=mx+c$ .

Slopes of perpendicular lines:

Given two lines $y=m_{1}x+c_1$ and $y=m_{2}x+c_2$ .

Observe that $m_1$ is the slope of the line $y=m_{1}x+c_1$ and $m_2$ is the slope of the line $y=m_{2}x+c_2$ .

Now, if $y=m_{1}x+c_1$ and $y=m_{2}x+c_2$ are perpendicular, then $m_1{m}_2=-1$ .

Calculation:

Find the slope of the line given in the graph:

Observe that the line given in the graph passes through the points $\left(0,0\right)$ and $\left(1,-3\right)$ .

Suppose that $m_1$ is the slope of the line.

Then, it follows that:

  $m_1=\frac{-3-0}{1-0}$ .

Find the value of $m_1$ :

  $\begin{array}{c}{m}_1=\frac{-3-0}{1-0}\\ m_1=\frac{-3}{1}\\ m_1=-3\end{array}$ .

Thus, the slope of the line:

  $m_1=-3$ .

Find the slope of the line perpendicular to the line given in the graph:

Suppose that $m_2$ is the slope of the line perpendicular to the line given in the graph.

Then, it follows that $m_1{m}_2=-1$ .

Substitute $m_1=-3$ in the equation $m_1{m}_2=-1$ :

  $-3m_2=-1$ .

Find the value of $m_2$ from $-3m_2=-1$ :

  $\begin{array}{c}-3m_2=-1\\ m_2=\frac{-1}{-3}\\ m_2=\frac{1}{3}\end{array}$ .

Identify the line with slope $\frac{1}{3}$from the given options:

Observe that the line $y=\frac{1}{3}x+5$ has the slope $\frac{1}{3}$ . Thus, the line $y=\frac{1}{3}x+5$ is the one that might be perpendicular to the line given in the graph.

Conclusion:

The line $y=\frac{1}{3}x+5$ seems to be the one that might be perpendicular to the line given in the graph.