Chapter 8.6, Problem 31E

a.

To determine

To find:the vector equation of the line.

Answer to Problem 31E

  $\left(x-11,y+4\right)=t\left(3,7\right)$

Explanation of Solution

Given information:

Line passes through point $P$ and is parallel to $\overrightarrow{a}$ .

  $\begin{array}{l}P=\left(11,-4\right)\\ \overrightarrow{a}=\left(3,7\right)\end{array}$

Calculation:

The vector equation of the line can be evaluated as follows:

  $\begin{array}{c}\left(x-x_1,y-y_1\right)=t\left(a_1,a_2\right)\end{array}$

  (x−11,y−(−4))=t(3,7)                      [Put components of P and a→]

  (x−7,y+4)=t(3,7)

Thus, the vector equation of line is $\left(x-11,y+4\right)=t\left(3,7\right)$ .

b.

To determine

To find:the parametric equation for the line.

Answer to Problem 31E

  $\begin{array}{l}x=3t+11\\ y=7t-4\end{array}$

Explanation of Solution

Given information:

The vector equation of a line evaluated in part a. is shown below.

  $\left(x-11,y+4\right)=t\left(3,7\right)$

Calculation:

The parametric equation for the lineare evaluated as follows:

  $\begin{array}{c}x-11=3t\\ x=11+3t\\ \\ y+4=7t\\ y=7t-4\end{array}$

Thus, the parametric equation of line is $x=3t+11\text{ and }y=7t-4$ .

c.

To determine

To find:the equation of line in slope intercept form by using the parametric equations.

Answer to Problem 31E

  $y=\frac{7}{3}x-\frac{89}{3}$

Explanation of Solution

Given information:

The parametric equations of line areas follows:

  $\begin{array}{l}x=3t+11\text{ and }\\ y=7t-4\end{array}$

Calculation:

Evaluate the value of $t$ in terms of $x$ as shown below.

  $\begin{array}{c}x=3t+11\\ x-11=3t\\ \frac{1}{3}x-\frac{11}{3}=t\\ or\\ t=\frac{1}{3}x-\frac{11}{3}\end{array}$

Substitute the value of $t$ in the parametric equation for $y$ and simply it as shown below.

  $\begin{array}{c}y=7t-4\\ =7\left(\frac{1}{3}x-\frac{11}{3}\right)-4\\ =\frac{7}{3}x-\frac{77}{3}-4\\ =\frac{7}{3}x-\frac{89}{3}\end{array}$

Thus, the slope intercept form of the line equation is $y=\frac{7}{3}x-\frac{89}{3}$ .