Chapter 1.5, Problem 29E

a.

To determine

To find: the value of $k$ .

Answer to Problem 29E

  $4$

Explanation of Solution

Given information:

Line equation $kx-7y+10=0$ is parallel to line $m$

Equation for line $m$ is $8x-14y+3=0$ .

Calculation:

Let the line equation $kx-7y+10=0$ be $l$ .

Rearrange the equation of line $m$ in the slope intercept form and determine the slope as shown below.

  $\begin{array}{c}kx-7y+10=0\\ 7y=kx+10\\ y=\frac{k}{7}x+\frac{10}{7}\\ \text{Compare with }y=mx+b\\ \therefore m=\frac{k}{7}\end{array}$

Thus, the slope of line $l$ is $\frac{k}{7}$ .

Rearrange the equation of line $m$ in the slope intercept form and determine the slope as shown below.

  $\begin{array}{c}8x-14y+3=0\\ 14y=8x+3\\ y=\frac{8}{14}x+\frac{3}{14}\\ y=\frac{4}{7}x+\frac{3}{14}\\ \text{Compare with }y=mx+b\\ \therefore m=\frac{4}{7}\end{array}$

Thus, the slope of line $m$ is $\frac{4}{7}$ .

Since the slope of parallel lines is same. Thus, we can equate the slopes of both the lines as follows:

  $\begin{array}{l}\parallel m_l=m_m\\ \frac{k}{7}=\frac{4}{7}\left[\text{Substituting the values of slopes}\right]\\ k=\frac{4}{7}\left(7\right)\\ k=4\end{array}$

Thus, the value of $k$ is 4.

b.

To determine

To find: the value of $k$ .

Answer to Problem 29E

  $-\frac{49}{4}$

Explanation of Solution

Given information:

Let the line equation $kx-7y+10=0$ be $l$ .

Line equation $l$ is perpendicular to line $m$

Equation for line $m$ is $8x-14y+3=0$ .

Slope of line $m$ = $\frac{4}{7}$

Slope of line $l$ = $\frac{k}{7}$

Calculation:

Since the slope of perpendicular line is opposite reciprocal of the other line. Thus, we can equate the slopes of both the lines as follows:

  $\begin{array}{l}\perp m_l=-\frac{1}{m_m}\\ \frac{k}{7}=-\frac{1}{\left(\frac{4}{7}\right)}\left[\text{Substituting value of }m\text{ for line }l\right]\\ \frac{k}{7}=-\frac{7}{4}\\ k=-\frac{7}{4}\left(7\right)\\ k=-\frac{49}{7}\end{array}$

Thus, the value of $k$ is $-\frac{49}{4}$ .