Chapter 1.3, Problem 79E

a.

To determine

Find equation of parallel line.

Answer to Problem 79E

  $\boxed{y=x+4.3}$

Explanation of Solution

Given information:

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

  $x-y=4$

  $\left(2.5,6.8\right)$

Calculation:

Consider the given points on line.

  $x-y=4$

  $\left(2.5,6.8\right)$

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 slope intercept form ,

  $y=mx+c$ ,

  $x-y=4$

  $\begin{array}{l}x-y=4\\ y=x-4\\ m=1\end{array}$

A line parallel to given line will have slope equal to the given line and it will pass through the given point $\left(2.5,6.8\right)$

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

Since line passes through the points $\left(2.5,6.8\right)$ .

  $\begin{array}{l}y=x+c\\ 6.8=2.5+c\\ c=4.3\end{array}$

So the line will be,

  $y=x+4.3$

Hence, the line through point $\left(2.5,6.8\right)$ that is parallel to the $x-y=4$ has equation in slope intercept form $y=mx+c$ $\boxed{y=x+4.3}$

b.

To determine

Find equation of perpendicular line.

Answer to Problem 79E

  $\boxed{y=-x+9.3}$

Explanation of Solution

Given information:

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

  $x-y=4$

  $\left(2.5,6.8\right)$

Calculation:

Consider the given points on line.

  $x-y=4$

  $\left(2.5,6.8\right)$

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 slope intercept form ,

  $y=mx+c$ ,

  $x-y=4$

  $\begin{array}{l}x-y=4\\ y=x-4\\ m=1\end{array}$

A line perpendicular to given line will have slope equal to the given line and it will pass through the given point $\left(2.5,6.8\right)$

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

Since line passes through the points $\left(2.5,6.8\right)$ .

  $\begin{array}{l}y=-x+b\\ 6.8=-2.5+c\\ c=9.3\\ y=-x+9.3\end{array}$

So the line will be,

  $y=-x+9.3$

Hence, the line through point $\left(2.5,6.8\right)$ that is parallel to the $x-y=4$ has equation in slope intercept form $y=mx+c$ $\boxed{y=-x+9.3}$