Chapter 0, Problem 49RE

To determine

To find: The parametric equation of line through the endpoints $\left(-2,5\right)$ and $\left(4,3\right)$ .

Answer to Problem 49RE

The parametric equation for the line through the endpoints $\left(-2,5\right)$ and $\left(4,3\right)$ is $x=t$ and $y=\frac{13-t}{3}$ .

Explanation of Solution

Given: the endpoints $\left(-2,5\right)$ and $\left(4,3\right)$ .

Concept Used:

Formula of Two-point slope-form of a line:

  $y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right)$

Calculation:

the endpoints $\left(-2,5\right)$ and $\left(4,3\right)$ .

Now,

  $\begin{array}{l}{x}_1=-2,y_1=5\\ x_2=4,y_2=3\end{array}$

Substituting the value in the formula:

  $\begin{array}{c}y-5=\frac{3-5}{4+2}\left(x-\left(-2\right)\right)\\ y-5=\frac{-2}{6}\left(x+2\right)\\ y-5=\frac{-1}{3}\left(x+2\right)\\ 3y-15=-x-2\\ 3y+x=-2+15\\ x+3y=13\end{array}$

Therefore, Equation of line is $x+3y=13$ .

Now in order to obtain the parametric equation, let $x=t$

Thus,

  $\begin{array}{c}x+3y=13\\ t+3y=13\left(\text{put},x=t\right)\\ y=\frac{13-t}{3}\end{array}$

Conclusion:

The parametric equation for the line through the endpoints $\left(-2,5\right)$ and $\left(4,3\right)$ is $x=t$ and $y=\frac{13-t}{3}$ .