Parameterization of Lines Examples

Parameterize the line

y = 3x + 5.

Since we're given the point-slope form, the first step is already done. The parameterization is 

x(t) = t 

y(t) = 3t + 5.

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

The point-slope equation of this line is

The parameterization is

x = t

Parametrize the line that goes through the points (2, 3) and (7, 9). 

The line looks like this:

Since we like going from left to right, put t = 0 at the point (2, 3). Since t = 1 is a nice number as well, put t = 1 at the point (7, 9).

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

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

First deal with x and forget y ever existed. Write an equation expressing x as a line in terms of t.

When t = 0 we have x = 2, and when t = 1 we 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.

Now forget about x. Write an equation expressing y as a line in terms of t.
When t = 0 we have y = 3 and when t = 1 we have y = 9.

The intercept is 3 and the slope is

The line expressing y in terms of t is y(t) = 3 + 6t.

The final parametrization is

x(t) = 2 + 5t
y(t) = 3 + 6t

To check that this parametrization is correct we need to make sure that its graph goes through the points (2, 3) and (7 ,9). When t = 0 we have

x = 2 + 5(0) = 2

y = 3 + 6(0) = 3

so the graph goes through (2,3). When t = 1 we have

x = 2 + 5(1) = 7

y = 3 + 6(1) = 9

so the graph goes through the point (7, 9). That means we found a correct parameterization, and we've made a hot dog with a bun and relish.

Parametrize the line that goes through the points (2, 3) and (7, 9). The parameterization should be at (7, 9) when t = 0 and should draw the line from right to left.

We're told that t = 0 should be (7, 9). We may as well put t = 1 at (2, 3) since that's a reasonable number.

Look at x 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 -5 (which makes sense, since x is decreasing as time passes). The equation is

x(t) = 7 – 5t.

Now look at y and t. When t = 0 we have y = 9 and when t = 1 we have y = 3.

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

The final parametrization is

x(t) = 7 – 5t

y(t) = 9 – 6t.

Parametrize the line that goes through the points (2, 3) and (7, 9) so that it takes 3 steps to travel from one point to the other. It's a chili dog.

Since the problem doesn't specify a direction we can 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 x = 7. 

The intercept is 2 and the slope is  

The equation for x is

Now look at y and t. When t = 0 we have y = 3. When t = 3 we have y = 9. 

The intercept is 3 and the slope is 

The equation for y is y = 3 + 2t.

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

Parametrize the line segment that connects the points (2, 3) 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) = 2 + 5t
y(t) = 3 + 6t.

We know that when t = 0 we're at the point (2, 3) and when t = 1 we're at the point (7, 9). If we restrict the parameter so that 0 ≤ t ≤ 1 then we find only the line segment that lies between those two points.