Parameterization Of A Line Segment

Points, Vectors, and Functions

Intro
Topics

Examples
Magnitude
Direction
Scaling Vectors
Unit Vectors
Sketching Vector Functions
Graphing Parametric Equations
Points on Graphs of Parametric Equations
Parametrizations of the Unit Circle
Parameterization of Lines
Polar Coordinates
Simple Polar Inequalities
Switching Coordinates
Translating Equations and Inequalities betwixt Coordinate Systems
Graphing Polar Functions
Rules of Graphing We Practice (or Don't) Have
Bounds on Theta
Intersections of Polar Functions

Exercises
Quizzes
Terms
Handouts
Best of the Web
Table of Contents

Parameterization of Lines Examples

Example 1

Parameterize the line

y = iii10 + 5.

Since we're given the signal-slope form, the first pace is already washed. The parameterization is

ten(t) = t

y(t) = 3t + 5.

Case 2

Parameterize the line that passes through the points (0, 1) and (4, 0).

The betoken-slope equation of this line is

The parameterization is

x = t

Example 3

Parametrize the line that goes through the points (ii, 3) and (7, 9).

The line looks like this:

Since nosotros like going from left to right, put t = 0 at the betoken (2, three). Since t = 1 is a nice number also, put t = ane at the point (7, 9).

We'll terminate with a parametrization that takes one time footstep to travel from one point to the other.

We're going to look at x and y separately. We'll terminate with a linear equation expressing ten in terms of t and a linear equation expressing y in terms of t.

First deal with 10 and forget y e'er existed. Write an equation expressing 10 as a line in terms of t.

When t = 0 we accept x = 2, and when t = one nosotros have x = 7.

The intercept of this line is 2 and its slope is

The line expressing x in terms of t is
x(t) = 2 + 5t.

At present forget about 10. Write an equation expressing y every bit a line in terms of t.
When t = 0 nosotros have y = three and when t = i we have y = 9.

The intercept is 3 and the gradient is

The line expressing y in terms of t is y(t) = 3 + vit.

The last parametrization is

10(t) = two + 5t
y(t) = 3 + 6t

To check that this parametrization is correct we need to make certain that its graph goes through the points (2, 3) and (seven ,9). When t = 0 we take

x = 2 + 5(0) = ii

y = iii + 6(0) = iii

then the graph goes through (ii,three). When t = ane we have

10 = 2 + 5(1) = 7

y = 3 + 6(1) = 9

so the graph goes through the point (seven, 9). That ways we establish a correct parameterization, and we've made a hot dog with a bun and relish.

Example 4

Parametrize the line that goes through the points (ii, 3) and (7, 9). The parameterization should be at (vii, ix) when t = 0 and should draw the line from correct to left.

We're told that t = 0 should be (7, ix). We may likewise put t = i at (2, three) since that's a reasonable number.

Expect at ten and t first. When t = 0 we have x = 7 and when t = 1 we have x = 2.

The intercept is 7 and the slope is -five (which makes sense, since x is decreasing every bit time passes). The equation is

x(t) = vii – fivet.

Now expect at y and t. When t = 0 nosotros take y = 9 and when t = 1 nosotros have y = 3.

The intercept is 9 and the gradient is -half dozen (which makes sense, since y is also decreasing as time passes). The equation is
y = 9 – 6t.

The final parametrization is

10(t) = 7 – 5t

y(t) = 9 – 6t.

Case five

Parametrize the line that goes through the points (2, three) and (7, ix) so that it takes 3 steps to travel from ane point to the other. It's a chili dog.

Since the problem doesn't specify a direction nosotros tin go from left to right. Put t = 0 at one point and t = 3 at the other.

First look at x and t. When t = 0 we have x = 2. When t = 3 we have ten = 7.

The intercept is 2 and the gradient is

The equation for x is

Now wait at y and t. When t = 0 we have y = 3. When t = 3 we have y = nine.

The intercept is three and the slope is

The equation for y is y = 3 + twot.

We haven't worried much about bounds for t because if we want the whole line, we want to have -∞ < t < ∞. We demand to give bounds for t if nosotros only want a piece of the line.

Example six

Parametrize the line segment that connects the points (two, three) and (7, 9). It's a miniature hot dog with relish.

This is like our first example, but it's asking for a line segment instead of the whole line. Take the parameterization we got earlier:

x(t) = two + 5t
y(t) = iii + 6t.

We know that when t = 0 we're at the point (ii, three) and when t = 1 we're at the bespeak (7, 9). If we restrict the parameter then that 0 ≤ t ≤ 1 then nosotros find simply the line segment that lies between those two points.