We are given parametric equations for a line as follows:

$x = t - 8$ $y = \frac{t}{4} - 3$

To find the rectangular equation (i.e., the equation in terms of $x$ and $y$):

1. First, solve for $t$ from the equation for $x$: $t = x + 8$

2. Now substitute this expression for $t$ into the equation for $y$: $y = \frac{x + 8}{4} - 3$

3. Simplify the equation: $y = \frac{x + 8}{4} - 3 = \frac{x}{4} + 2 - 3 = \frac{x}{4} - 1$

Thus, the rectangular equation of the line is: $y = \frac{x}{4} - 1$

Graphing the Line and Showing Orientation

• The equation $y = \frac{x}{4} - 1$ represents a line with a slope of $\frac{1}{4}$ and a y-intercept of $-1$.
• To indicate the direction (orientation), as $t$ increases, $x = t - 8$ and $y = \frac{t}{4} - 3$ both increase. This means the line is traced in the direction of increasing $x$ and $y$ as $t$ increases.

If you'd like, I can generate a graph of this line along with its orientation. Would you like that?

Here are five related questions for further practice:

1. How would the rectangular equation change if $x = 2t - 5$ and $y = \frac{t}{3} + 1$?
2. What happens to the slope of the line if the parametric equations are $x = t - 4$ and $y = 2t - 3$?
3. Can you convert the parametric equations $x = t^2$ and $y = 2t + 1$ into a rectangular equation?
4. What would be the parametric equations for a line with a slope of 2 and passing through the point (3, 1)?
5. How can you verify that a parametric equation represents a linear equation?

Tip:

When solving parametric equations, always solve for $t$ in terms of $x$ or $y$ first to make substitution easier.