Chapter 6.3, Problem 59E

Solution

To find:whether the given statement is True or False.

True

Given information:

The parametric equations: $x_1=t-1$ , $y_1=3t+1$ and $x_2=\left(2/3\right)t-4/3$ , $y_2=2t$

Calculation:

Eliminate $t$:

Consider the first set of parametric equations: $x_1=t-1$ , $y_1=3t+1$ ;

Solve for $t$ from the equation: $x_1=t-1$

  $\begin{array}{l}{x}_1=t-1\\ t=x_1+1\end{array}$

Substitute the value of $t$ , in the equation $y_1=3t+1$ :

  $\begin{array}{l}{y}_1=3t+1\\ y_1=3\left(x_1+1\right)+1\\ y_1=3x_1+4\end{array}$

The rectangular equation is $y_1=3x_1+4$ .

Consider the first set of parametric equations: $x_2=\left(2/3\right)t-4/3$ , $y_2=2t$ ;

Solve for $t$ from the equation: $x_2=\left(2/3\right)t-4/3$

  $\begin{array}{c}{x}_2=\left(2/3\right)t-4/3\\ \left(2/3\right)t=x_2+4/3\\ t=\left(3/2\right)x_2+2\end{array}$

Substitute the value of $t$ , in the equation $y_2=2t$ :

  $\begin{array}{c}{y}_2=2t\\ y_2=2\left(\left(3/2\right)x_2+2\right)\\ =3x_2+4\end{array}$

The rectangular equation is $y_2=3x_2+4$ .

Therefore, both sets of parametric equations correspond to the same rectangular equation.

Thus, given statement is True.