Chapter 1.3, Problem 78E

a.

To determine

Find equation of parallel line.

Answer to Problem 78E

  $\boxed{x=3}$

Explanation of Solution

Given information:

Write equations of the lines through the given point parallel to given line.

  $x-4=0$

  $\left(3,-2\right)$

Calculation:

Consider the given points on line.

  $x-4=0$

  $\left(3,-2\right)$

The equation of line with slope $m$ and given point on the line is,

  $m=\frac{y-y_1}{x-x_1}$

If the slopes of two non vertical lines are equal then lines are parallel.

  $m_1=m_2$

If the slopes of two non vertical lines are negative reciprocal of each other then lines are perpendicular.

  $m_1=-\frac{1}{m_2}$

Now compare given line with standard slope intercept form ,

  $y=mx+c$ ,

  $x-4=0$

  $\begin{array}{l}x-4=0\\ x=4\\ m=undefined\end{array}$

A line parallel to given line will have slope equal to the given line and it will pass through the given point $\left(3,-2\right)$

  $\begin{array}{l}y=m_{1}x+c\\ m_1=m\\ m_1=0\\ x=c\end{array}$

Since line passes through the points $\left(3,-2\right)$ t

  $\begin{array}{l}x=c\\ 3=c\\ so\\ \boxed{x=3}\end{array}$

Hence, the line through point $\left(3,-2\right)$ that is parallel to the $x-4=0$ has equation in slope intercept form $y=mx+c$

  $\boxed{x=3}$

b.

To determine

Find equation of perpendicular line.

Answer to Problem 78E

  $\boxed{y=-2}$

Explanation of Solution

Given information:

Write equations of the lines through the given point parallel to given line.

  $x-4=0$

  $\left(3,-2\right)$

Calculation:

Consider the given points on line.

  $x-4=0$

  $\left(3,-2\right)$

The equation of line with slope $m$ and given point on the line is,

  $m=\frac{y-y_1}{x-x_1}$

If the slopes of two non vertical lines are equal then lines are parallel.

  $m_1=m_2$

If the slopes of two non vertical lines are negative reciprocal of each other then lines are perpendicular.

  $m_1=-\frac{1}{m_2}$

According to the above perpendicular line property to the given line

  $x-4=0$ must have slope

  $\begin{array}{l}y=m_{2}x+c\\ m_2=-\frac{1}{m}\\ m_2=0\\ y=c\\ -2=c\\ y=-2\end{array}$

Hence, the line through point $\left(3,-2\right)$ that is perpendicular to the $x-4=0$ has equation in slope intercept form $y=mx+c$

  $\boxed{y=-2}$