Chapter 1.3, Problem 74E

a.

To determine

Find equation of parallel line.

Answer to Problem 74E

  $\boxed{y=-x-1}$

Explanation of Solution

Given information:

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

  $x+y=7$

  $\left(-3,2\right)$

Calculation:

Consider the given points on line.

  $x+y=7$

  $\left(-3,2\right)$

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 equation of line,

  $x+y=7$

Subtract $x$ from both sides,

  $\begin{array}{l}x+y-x=7-x\\ y=-x+7\end{array}$

New equation $\boxed{y=-x+7}$ is in the form of lope intercept form $y=mx+c$

  $\boxed{m=-1}$

According to above parallel line property any line parallel to considered line $x+y=7$ must have slope $m=-1$ .

By using point slope form equation,

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-2=-1\left(x-\left(-3\right)\right)\\ y-2=-1\left(x+3\right)\\ y-2=-x-3\end{array}$

Now add $2$ both sides in above equation,

  $\begin{array}{l}y-2+2=-x-3+2\\ y=-x-1\end{array}$

Hence, the line through point $\left(-3,2\right)$ that is parallel to the $x+y=7$ has equation in slope intercept form $y=mx+c$ is $\boxed{y=-x-1}$

b.

To determine

Find equation of perpendicular line.

Answer to Problem 74E

  $\boxed{y=x+5}$

Explanation of Solution

Given information:

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

  $x+y=7$

  $\left(-3,2\right)$

Calculation:

Consider the given points on line.

  $x+y=7$

  $\left(-3,2\right)$

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 equation of line,

  $x+y=7$

Subtract $x$ from both sides,

  $\begin{array}{l}x+y-x=7-x\\ y=-x+7\end{array}$

New equation $\boxed{y=-x+7}$ is in the form of lope intercept form $y=mx+c$

  $\boxed{m=-1}$

Now $m_1=-\frac{1}{m_2}$

  $m=-\frac{1}{-1}=1$

According to above perpendicular line property any line parallel to considered line $x+y=7$ must have slope $m=-\frac{1}{-1}=1$ .

By using point slope form equation,

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-2=1\left(x-\left(-3\right)\right)\\ y-2=1\left(x+3\right)\\ y-2=x+3\end{array}$

Now add $2$ both sides in above equation,

  $\begin{array}{l}y-2+2=x+3+2\\ y=x+5\end{array}$

Hence, the line through point $\left(-3,2\right)$ that is perpendicular to the $x+y=7$ has equation in slope intercept form $y=mx+c$ is $\boxed{y=x+5}$