Chapter 13.5, Problem 50E

To determine

To graph: The equation of line where parametric equations are $x=4t\text{ and }y=3+2t$ .

Explanation of Solution

Given information:

The parametric equations are $x=4t\text{ and }y=3+2t$ .

Graph:

Consider the parametric equations are $x=4t\text{ and }y=3+2t$ .

Compute the value of t from both the equations.

First take the equation $x=4t$ ,

  $\begin{array}{c}x=4t\\ t=\frac{x}{4}\end{array}$

Next take the equation $y=3+2t$ ,

  $\begin{array}{c}y=3+2t\\ 2t=y-3\\ t=\frac{y-3}{2}\end{array}$

Now, equate both the values of t ,

  $\begin{array}{c}\frac{y-3}{2}=\frac{x}{4}\\ 4y-12=2x\\ -2x+4y=12\\ -x+2y=6\end{array}$

Write it in slope intercept form,

  $\begin{array}{c}-x+2y=6\\ 2y=x+6\\ y=\frac{1}{2}x+3\end{array}$

Sketch the equation $y=\frac{1}{2}x+3$ in the coordinate plane.

  

Interpretation:

The equation $y=\frac{1}{2}x+3$ is a straight line that has x -intercept as $-6$ and y -intercept as 3.