Chapter 4.3, Problem 15E

To determine

which of the lines are parallel or perpendicular.

Answer to Problem 15E

The lines $a,bandc$ are not parallel to each other.

The lines $bandc$ are perpendicular to each other.

Explanation of Solution

Given:

Line a passes through (-2,1) and (0,3)

Line b passes through (4,1) and (6,4)

Line c passes through (1,3) and (4,1)

Calculation:

Consider the lines $a,b,andc$ such that the coordinates forline $aare\left(-2,1\right)and\left(0,3\right)$ , the coordinates for line $bare\left(4,1\right)and\left(6,4\right)$ and the coordinates for line $care\left(1,3\right)and\left(4,1\right)$ .

The objective is to determine which lines are parallel or perpendicular.

To determine whether the lines are parallel or perpendicular, find the slopes of lines $a,b,andc$ .

The slope of a line is the ratio of difference of $ycoordinatetoxcoordinate$ .

Therefore, the slope of line $a$ is,

  $\begin{array}{l}{m}_a=\frac{1-3}{-2-0}\\ =\frac{-2}{-2}\\ =1\end{array}$

The slope of line $b$ is,

  $\begin{array}{l}{m}_b=\frac{1-4}{4-6}\\ =\frac{-3}{-2}\\ =\frac{3}{2}\end{array}$

Similarly, the slope ofline $c$ is,

  $\begin{array}{l}{m}_c=\frac{3-1}{1-4}\\ =\frac{2}{-3}\\ =-\frac{2}{3}\end{array}$

Since the slopes of line $a,bandc$ are different, therefore, lines $a,bandc$ are not parallel to each other.

But, the slope of lines $bandc$ is negative inverse of each other.

That is,

  $\begin{array}{l}{m}_c=\frac{-6-\left(-1\right)}{-3-0}\\ =\frac{-5}{-3}\\ =\frac{5}{3}\\ a,bandc\\ m_b{m}_c=\frac{3}{2}\left(-\frac{2}{3}\right)\\ m_b{m}_c=-1\end{array}$

Therefore, the lines $bandc$areperpendicular to each other.

Conclusion:

The lines $a,bandc$ are not parallel to each other.

The lines $bandc$ are perpendicular to each other.