Chapter 8.6, Problem 16E

To determine

To find: a vector equation and parametric equations of a line passing through point $P$ and parallel to $\overrightarrow{a}$ .

Answer to Problem 16E

Vector equation:

  $\left(x-1,y-0\right)=t\left(-2,-4\right)$

Parametric equations:

  $\begin{array}{c}x=1-2t\\ y=-4t\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}P\left(1,0\right)\\ \overrightarrow{a}=\left(-2,-4\right)\end{array}$

Calculation:

Substitute the values in the vector equation given below.

  $\begin{array}{c}\left(x-x_1,y-y_1\right)=t\left(a_1,a_2\right)\\ \left(x-1,y-0\right)=t\left(-2,-4\right)\end{array}$

Thus the vector equation is $\left(x-1,y-0\right)=t\left(-2,-4\right)$ .

Substitute the required values $\left(x_1,y_1\right)=\left(1,0\right)$ and $\left(a_1,a_2\right)=\left(-2,-4\right)$ in the general form of parametric equations shown below.

  $\begin{array}{c}x=x_1+t{a}_1\\ =1+t\left(-2\right)\\ =1-2t\end{array}$

  $\begin{array}{c}y=y_1+t{a}_2\\ =0+t\left(-4\right)\\ =-4t\end{array}$

Thus the parametric equations for a line parallel to the given line and passing through the given point are shown below.

  $\begin{array}{c}x=1-2t\\ y=-4t\end{array}$