91Ó°ÊÓ

The direction vector is a crucial concept when working with parametric equations in geometry. It represents the direction in which a line extends. To find this, you subtract the coordinates of the initial point from the terminal point.
In our problem, we determine the direction vector by taking the points given:

• Point A: (1, -3)
• Point B: (4, 0)

Subtract the coordinates of Point A from Point B:

• Increased by 3 in the x-direction: 4 - 1 = 3
• Increased by 3 in the y-direction: 0 - (-3) = 3

Thus, our direction vector is (3, 3).
A direction vector like (3, 3) indicates that as you move along the line, you ascend by three units on both x and y axes for each step of the parameter 't'. This direction is consistent throughout the line and provides a simple way to express the line's trajectory.

The standard form of a line's equation is an algebraic expression that presents the line in a specific format: \[ Ax + By = C \] where A, B, and C are integers, and A should be a positive number.
In our exercise, we first set out to find this form by eliminating the parameter from the parametric equations. Starting with the parametric equations:

• \( x(t) = 1 + 3t \)
• \( y(t) = -3 + 3t \)

We found:

• Solve for 't' from the equation of x: \( t = \frac{x - 1}{3} \)

Then, substitute into y:

• \( y = -3 + 3\left(\frac{x-1}{3}\right) \)
• \( y = x - 4 \)

This can be arranged to the standard form:

• \( x - y = 4 \)

This effectively removes any parametric terms and presents the line in a clear, integer-based way. The coefficients resemble the relationship between x and y, offering a straightforward way to check line parallelism and calculate intersections, among other tasks.

Eliminating the parameter in a set of parametric equations means rewriting the equations so that the variable that represents the parameter is no longer present. This process allows you to convert parametric equations into more conventional forms like slope-intercept or standard form.
Begin by isolating one parameter, typically t, in the parametric equations. From the exercise, starting with:

• \( x(t) = 1 + 3t \)

We solve for 't':

• \( t = \frac{x - 1}{3} \)

Next, substitute 't' in the equation for y:

• \( y(t) = -3 + 3t \)

This results in:

• \( y = -3 + 3\left(\frac{x-1}{3}\right) \)
• Hence, \( y = x - 4 \)

By eliminating 't', we've succinctly expressed the relationship between x and y, concluding the transition to a more familiar line equation. The goal of eliminating parameters is to showcase the functional relationship directly, which is easier for analytical tasks in algebraic contexts.