Chapter 1.3, Problem 76E

a.

To determine

Find equation of parallel line.

Answer to Problem 76E

  $\boxed{8y=-40x+41}$

Explanation of Solution

Given information:

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

  $5x+3y=0$

  $\left(\frac{7}{8},\frac{3}{4}\right)$

Calculation:

Consider the given points on line.

  $5x+3y=0$

  $\left(\frac{7}{8},\frac{3}{4}\right)$

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,

  $5x+3y=0$

Subtract $5x$ from both sides,

  $\begin{array}{l}5x+3y-5x=0-5x\\ 3y=-5x\\ \boxed{m=-5}\end{array}$

Now equation $\boxed{3y=-5x}$ is in the form of lope intercept form $y=mx+c$

  $\boxed{m=-5}$

According to above parallel line property any line parallel to considered line $5x+3y=0$ must have slope $\boxed{m=-5}$ .

By using point slope form equation,

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\frac{3}{4}=-5\left(x-\frac{7}{8}\right)\\ \frac{4y-3}{4}=-5\left(\frac{8x-7}{8}\right)\\ \frac{4y-3}{4}=\left(\frac{-40x+35}{8}\right)\\ 4y-3=\frac{-40x+35}{2}\\ \end{array}$

Now multiply $2$ both sides in above equation,

  $\begin{array}{l}\left(4y-3\right)\times 2=\frac{-40x+35}{2}\times 2\\ 8y-6=-40x+35\end{array}$

Now add $6$ both sides in above equation

  $\begin{array}{l}8y-6+6=-40x+35+6\\ 8y=-40x+41\end{array}$

Hence, the line through point $\left(\frac{7}{8},\frac{3}{4}\right)$ that is parallel to the $5x+3y=0$ has equation in slope intercept form $y=mx+c$

  $\boxed{8y=-40x+41}$

b.

To determine

Find equation of perpendicular line.

Answer to Problem 76E

  $\boxed{8y=40x-29}$

Explanation of Solution

Given information:

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

  $5x+3y=0$

  $\left(\frac{7}{8},\frac{3}{4}\right)$

Calculation:

Consider the given points on line.

  $5x+3y=0$

  $\left(\frac{7}{8},\frac{3}{4}\right)$

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

  $5x+3y=0$ must have slope $-\frac{1}{\left(-5\right)}$

So the line through point $\left(\frac{7}{8},\frac{3}{4}\right)$ that is perpendicular to the line $5x+3y=0$ must have equation,

By using point slope form equation $y-y_1=m\left(x-x_1\right)$

So $m=\frac{1}{5}$

  $\begin{array}{l}m=5\\ x_1=\frac{7}{8},y_1=\frac{3}{4}\end{array}$

  $\begin{array}{l}y-y_1=m\left(x-x_1\right)\\ y-\frac{3}{4}=5\left(x-\frac{7}{8}\right)\\ \frac{4y-3}{4}=5\left(\frac{8x-7}{8}\right)\\ \frac{4y-3}{4}=\left(\frac{40x-35}{8}\right)\\ \left(4y-3\right)=\frac{40x-35}{2}\end{array}$

Multiply both side by $2$

  $\begin{array}{l}\left(4y-3\right)\times 2=\frac{40x-35}{2}\times 2\\ 8y-6=40x-35\end{array}$

Add $6$ to both sides,

  $\begin{array}{l}8y-6=40x-35\\ 8y-6+6=40x-35+6\\ \boxed{8y=40x-29}\end{array}$

Hence, the line through point $\left(\frac{7}{8},\frac{3}{4}\right)$ that is perpendicular to the $5x+3y=0$ has equation in slope intercept form $y=mx+c$

  $\boxed{8y=40x-29}$