A parametric curve is a pair of functions

y = g(t)

where the two continuous functions define ordered pairs **(x,y)**.
The extent of the curve is dependent on the range one sets for **t**.

When looking at parametric equations, there are a couple of things one can get out of just the equations without even graphing them. In this discussion, the parametric equations that are explored are linear parametric equations or parametric equations of a line as they are more commonly called. If given two parametric equations

y = c + dt

what can be understood from them before graphing them. The first thing
to notice is that the slope of the line that these two parametric curves
define is equal to. This is illustrated in **Figure 1**
below.

The pink line above is the graph of the parametric equations

y = t

The values of **a** and **c** are zero, **b** and **d** are
1, and **t** was in the range from 0 to 10. From this figure, notice
that when **x=2**, **y=2 **and when **x=6**, **y=6**. This tells
you that the slope of the line must be 1, which satisfies the statement
above that the slope was the quotient of **d **and **b**. Another
piece of information that one can obtain from two parametric equations that
define a line, is the point at which the graph will start. In **Figure
1**, the values of **a** and **c** and the initial value of **t**
was zero and thus the point at which the graph started was **(0,0)**.
So one would think that these points have something to do with were the
graph starts and in fact they will define exactly where the graph will start.
If the value of **t** is zero, then the point at which the graph starts
will be **(a,c)**. **Figure 2** shows a graphical description of this.

From looking at this graph, one will notice that the graph does not start
at **(0,0)** therefore **a** and **c** must not equal zero. The
parametric equations for **Figure 2** are

y = 1 + t

for a range of **t** between 0 and 10. Since **t=0** is the first
time that becomes, this yields the start point for the graph to be at **(1,1)**,
which is consistent with the graph in **Figure 2**. Therefore, when the
initial time for **t** is zero, the starting point of the the graph will
be located at **(a,c)**. Where would you expect the graph to start when
the initial value of **t** is not zero? The answer to this comes right
out the what was just discussed. To find the starting point of the graph
of parametric equations of a line, one must only add the value of **t**
to the value of **a** and the value of **c**, and these two new values
will be the coordinates of the starting point. Or more simply, the starting
point of the graph of parametric equations of a line will be located at
**(t+a,t+c)**.

Now that we know how to determine what the slope and starting point will
be for parametric equations of a line, we will know try to solve the following
problem: Write parametric equations of a line that pass through (5,7) with
a slope of 3. You know that there can only be one line that is defined from
positive infinity to negative infinity that passes through (5,7) with a
slope of 3. However, parametric equations allows us to break this one line
down into smaller segments but the segments that we define must still pass
through (5,7) with a slope of 3. There are many different equations one
could write to satisfy this problem. All of the solutions listed here take
into account that **t** varied from 0 to 10.

**Figure 3** is the graph of the parametric equations

y = -2 + 3t

From this figure you can see that the graph (we are talking about the
pink line in the graph) starts at point **(a,c)** and has a slope of
3 and passes through the point (5,7). The green line is the line x=5 and
the aqua line is the line y=7. These lines were included to show that the
point (5,7) does in fact lie on the parametric equations of a line defined
for this graph. **Figure's 4** and **5** show two more examples of
paramtric equations of a line that have slope 3 and pass through the point
(5,7). The equations represented in the two graphs are

y = -14 + 3t

y = 7 + 3t

respectively.

The examples of parametric equations of a line given here are just three
of the many possibilities one can form. An easy way to form pairs of parametric
equations that fit the requirements of passing through (5,7) and have a
slope of three is to determine one of the equations first. Then solve this
equation to find out at what value of **t** the parametric equations
of a line will be located at the point on the line that was given as a requirement.
Then take this time and the fact thatmust equal the required
slope to find the other equation. Once a pair of equations is formed, you
can get many different variations of those equations by simply changing
the value of **t** to start and stop the graph of different times. All
of the graphs will be using the same pair of equations, but the line segment
that they define will be different for each different value of **t**.

Parametric equations of a line are only one family of parametric equations. Some of the other families of parametric equations create some really odd shapes. It is left up to the reader to explore the many other possibilities of aprametric curves.