Chapter 4.3, Problem 17E

To determine

which of the lines are parallel or perpendicular.

Answer to Problem 17E

The lines $aandb$ are parallel.

The line $c$ is perpendicular to both the lines $aandb$ .

Explanation of Solution

Given:

  $Linea:4x-3y=2$

  $Lineb:\text{ y}=\frac{4}{3}x+2$

  $Linec:\text{ 4y+}3x=4$

Formula Used:

  $y=mx+c$

Calculation:

Consider the following equations of line $a,b,andc$ :

  $Linea:4x-3y=2$

  $Lineb:\text{ y}=\frac{4}{3}x+2$

  $Linec:\text{ 4y+}3x=4$

The objective is to determine whether the lines are parallel or perpendicular.

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

To calculate the slope of lines, reduce them in the standard form of $y=mx+c$ , where $\left(x,y\right)$ is the coordinate of line, $c$ is constant value, and $m$ is slope of the line.

Line a:

  $\begin{array}{l}4x-3y=2\\ 3y=4x-2\\ \frac{3}{3}y=\frac{4}{3}x-\frac{2}{3}\\ y=\frac{4}{3}x-\frac{2}{3}\end{array}$

Compare it with the standard form,

Thus, $m_a=\frac{4}{3}$

Hence, slope of line $a=\frac{4}{3}$ .

Line b:

  $y=\frac{4}{3}x+2$

Compare it with the standard form.

Thus, $m_b=\frac{4}{3}$

Hence, slope of line $b=\frac{4}{3}$

Line c:

  $\begin{array}{l}4y+3x=4\\ 4y=-3x+4\\ \frac{4}{4}y=-\frac{3}{4}x+\frac{4}{4}\\ y=-\frac{3}{4}x+1\end{array}$

Compare it with the standard form.

Thus, $m_c=-\frac{3}{4}$ .

Hence, slope of line $c=-\frac{3}{4}$ .

Since the slope of all the lines $aandb$ are equal, therefore lines $aandb$are parallel.

And slope of line $c$ is negative inverse of slope of both the lines $aandb$ .

That is,

  $\begin{array}{l}{m}_a{m}_c=\frac{4}{3}\left(-\frac{3}{4}\right)\\ m_a{m}_c=-1\\ And\\ m_b{m}_c=\frac{4}{3}\left(-\frac{3}{4}\right)\\ m_b{m}_c=-1\end{array}$

Hence, the line $c$is perpendicular to both the lines $aandb$ .

Conclusion:

The lines $aandb$ are parallel.

The line $c$ is perpendicular to both the lines $aandb$ .