Question

In Exercises \(23-28,\) find a parametrization for the curve. the line segment with endpoints \((-1,3)\) and \((3,-2)\)

Short answer

The parametrization of the given line segment is \(r(t) = (-1 + 4t, 3 - 5t)\), for \(0 \leq t \leq 1\).

Step-by-step solution

Identify Endpoints

The line segment given has endpoints (-1,3) and (3,-2). Let's denote these points as \(A(-1, 3)\) and \(B(3, -2)\).

Compute Vector AB

Compute the vector AB which goes from point A to point B. We find this by subtracting the coordinates of A from B. \nSo, \(AB = B - A = (3 - (-1), -2 - 3) = (4, -5)\).

Parametrize the Line Segment

The line segment AB is represented by the parametric equation in vector form: \(r(t) = A + tAB\), where \(0 \leq t \leq 1\). This equation gives all the points from A to B as t varies from 0 to 1. Now substitute the values of A and AB in the equation. The parametric form would then be, \(r(t) = (-1,3) + t(4, -5)\), which simplifies further to \(r(t) = (-1 + 4t, 3 - 5t)\).

Key concepts

Parametric Equations

Understanding how to describe the path of a line segment involves the use of parametric equations. These equations express the coordinates of points on the line segment as functions of a parameter 't'. Instead of depicting a line as a single equation in two variables (x and y), parametric equations break this into two separate equations: one for x and one for y, each in terms of the parameter t.

For a line segment with endpoints A and B, a straightforward parametrization is to start at point A and move towards point B as the parameter t increases from 0 to 1. In the case of the given problem, the ends of the line segment are given as A(-1, 3) and B(3, -2). The parametric equations are then \( x(t) = -1 + 4t \) and \( y(t) = 3 - 5t \) where \( 0 \leq t \leq 1 \). As t varies from 0 to 1, these equations describe the x and y coordinates of every point on the segment from A to B.

Vector Arithmetic

In the given problem, we apply vector arithmetic to find the vector pointing from one endpoint of a line segment to the other. A vector representing the direction and distance from point A to B is denoted as \( \overrightarrow{AB} \), and it can be calculated by subtracting the coordinates of point A from point B's coordinates.

Using the example of the endpoints A(-1, 3) and B(3, -2), the corresponding vector \( \overrightarrow{AB} \) is computed as \( (3 - (-1), -2 - 3) = (4, -5) \). This vector indicates how to move from A to B: 4 units to the right from A and 5 units down. During parametrization, the vector is scaled by the parameter t. The parameter 't' determines the portion of the vector that's been covered, enabling us to trace the line segment as t goes from 0 to 1.

Coordinate Geometry

The principles of coordinate geometry are essential for solving the problem at hand. Coordinate geometry allows us to use algebra to describe and analyze geometric shapes. In the context of our problem, we're dealing with the simplest shape: a line segment in a two-dimensional plane. The endpoints of the segment give us specific coordinates which we use to determine the straight-line path between them.

The parametrization process explained in the previous sections takes these endpoints and translates them into a formula that can provide the location of any intermediate point on the line segment. Coordinate geometry makes it possible to understand that by combining the starting point with the vector direction scaled by parameter t, any point on the line segment can be precisely defined. Therefore, for the line segment with endpoints A(-1, 3) and B(3, -2), every point on the segment can be expressed as \( (-1 + 4t, 3 - 5t) \) where t is between 0 and 1. This approach allows us to seamlessly merge algebra with geometry, offering a clear mathematical framework to solve such geometric problems.