### Video Transcript

In this video, we will learn how to
find the equation of a straight line in parametric form using a point on the line
and the direction vector of the line. Letโs begin by recalling the vector
form of a straight line. The vector form of a straight line
passing through the point ๐ด and parallel to the direction vector ๐ is ๐ซ is equal
to ๐๐ plus ๐ก multiplied by ๐. This can be represented on the
two-dimensional coordinate plane as shown.

We recall that the position vector
of a point is the vector starting from the origin and ending at that point. The vector form of the equation of
a line describes each point on the line as its position vector ๐ซ. Each value of the parameter ๐ก
gives the position vector of one point on the line.

If we consider a line passing
through the point ๐ด with coordinates ๐ฅ sub zero, ๐ฆ sub zero and parallel to the
direction vector ๐ with components ๐ and ๐, then the vector form of the equation
of the line is given by ๐ซ is equal to ๐ฅ sub zero, ๐ฆ sub zero plus ๐ก multiplied
by ๐, ๐. Simplifying the right-hand side of
our equation, we get the components ๐ฅ sub zero plus ๐๐ก and ๐ฆ sub zero plus
๐๐ก. We can then write the position
vector on the left-hand side in terms of its ๐ฅ- and ๐ฆ-components. This leads us to the parametric
form of the equation of a straight line. The parametric form of the equation
of a line passing through the point ๐ด with coordinates ๐ฅ sub zero, ๐ฆ sub zero and
parallel to the direction vector ๐ is ๐ฅ is equal to ๐ฅ sub zero plus ๐๐ก, ๐ฆ is
equal to ๐ฆ sub zero plus ๐๐ก. In our first question, we will look
at an example of this in practice.

Straight line ๐ฟ passes through the
point ๐ with coordinates three, four and has a direction vector ๐ฎ equal to two,
negative five. Then the parametric equations of
line ๐ฟ are what.

We begin by recalling that the
parametric form of the equation of a line passing through the point ๐ฅ sub zero, ๐ฆ
sub zero and parallel to the direction vector ๐, ๐ is ๐ฅ is equal to ๐ฅ sub zero
plus ๐๐ก and ๐ฆ is equal to ๐ฆ sub zero plus ๐๐ก. We are told in the question that
the straight line ๐ฟ passes through the point with coordinates three, four. This means that our value of ๐ฅ sub
zero is three and ๐ฆ sub zero is equal to four. We are also given a direction
vector ๐ฎ such that ๐ is equal to two and ๐ is negative five.

Substituting in our values of ๐ฅ
sub zero and ๐, we get ๐ฅ is equal to three plus two ๐ก. And substituting the values of ๐ฆ
sub zero and ๐, we get ๐ฆ is equal to four minus five ๐ก. It is important to note that we
could replace the letter ๐ก with any other letter as the parameter. For example, ๐ฅ is equal to three
plus two ๐ and ๐ฆ is equal to four minus five ๐ is also a valid solution. We can therefore conclude that
these are the parametric equations of line ๐ฟ.

We will now look at another example
looking at the process of converting the vector form to the parametric form.

The vector equation of a straight
line is given by ๐ซ of ๐ก is equal to ๐ก multiplied by five, two plus negative one,
three. Which of the following pairs of
parametric equations represents this straight line? Is it (A) ๐ฅ is equal to five ๐ก
plus two, ๐ฆ is equal to negative ๐ก plus three? (B) ๐ฅ is equal to five ๐ก minus
one, ๐ฆ is equal to two ๐ก plus three. (C) ๐ฅ is equal to three ๐ก plus
two, ๐ฆ is equal to negative ๐ก plus five. (D) ๐ฅ is equal to negative ๐ก plus
five, ๐ฆ is equal to three ๐ก plus two. Or (E) ๐ฅ is equal to two ๐ก plus
three, ๐ฆ is equal to five ๐ก minus one.

We begin this question by recalling
that the vector form of the equation of a line is ๐ซ is equal to ๐ฅ sub zero, ๐ฆ sub
zero plus ๐ก multiplied by ๐, ๐, where ๐ฅ sub zero, ๐ฆ sub zero is the position
vector of a point on the line and ๐, ๐ is a direction vector of the line.

Comparing this to the equation
given, we see that the direction vector for our line is five, two. The position vector when ๐ก equals
zero is negative one, three. This means that our line passes
through the point one, three and is parallel to the direction vector five, two.

Next, we recall that the parametric
form of the equation of a line is ๐ฅ equals ๐ฅ sub zero plus ๐๐ก and ๐ฆ equals ๐ฆ
sub zero plus ๐๐ก. Substituting in the values from
this question, we have ๐ฅ is equal to negative one plus five ๐ก and ๐ฆ is equal to
three plus two ๐ก. Noting the way the equations have
been written in the five options, we have ๐ฅ is equal to five ๐ก minus one and ๐ฆ is
equal to two ๐ก plus three. The correct answer is option
(B).

In our next example, we will apply
the definition for the parametric form to obtain the direction vector. The direction vector of the
straight line whose parametric equations are ๐ฅ equals two and ๐ฆ equals negative
two ๐ plus four is what.

We begin by recalling that the
parametric form of the equation of a line passing through the point with coordinates
๐ฅ sub zero, ๐ฆ sub zero and parallel to the direction vector with components ๐, ๐
is ๐ฅ is equal to ๐ฅ sub zero plus ๐๐ก and ๐ฆ is equal to ๐ฆ sub zero plus
๐๐ก. We are given the parametric
equations ๐ฅ equals two and ๐ฆ equals negative two ๐ plus four.

Noting that the parameter here is
๐, we can rewrite the general equations as shown. Comparing terms, we see that ๐ฅ sub
zero is equal to two. ๐ is equal to zero, as there is no
๐ term in our first parametric equation. We also have ๐ฆ sub zero is equal
to four and ๐ is equal to negative two. This means that our line passes
through the point two, four. It is also parallel to the
direction vector zero, negative two. We can therefore conclude that the
direction vector of the straight line whose parametric equations are ๐ฅ equals two
and ๐ฆ equals negative two ๐ plus four is zero, negative two.

Whilst it is not required in this
question, we could use this information to write the vector equation of the
line. The position vector ๐ซ is equal to
two, four plus ๐ multiplied by zero, negative two. As we have seen in this question,
instead of giving the direction vector directly, a problem may provide this
indirectly. In fact, the direction vector of a
line may be given indirectly in three possible ways: firstly, by providing two
points that lie on the line; secondly, by providing the angle between the line and
the positive ๐ฅ-axis; and thirdly, by providing the slope or gradient of the
line. We will now consider a couple of
these scenarios.

Find the parametric equations of
the straight line that makes an angle of 135 degrees with the positive ๐ฅ-axis and
passes through the point one, negative 15. Is it (A) ๐ฅ is equal to one plus
๐, ๐ฆ is equal to negative 15 minus ๐? (B) ๐ฅ is equal to one plus ๐, ๐ฆ
is equal to one minus 15๐. Is it (C) ๐ฅ is equal to negative
15 minus ๐, ๐ฆ is equal to one plus ๐? Or (D) ๐ฅ is equal to one, ๐ฆ is
equal to negative 15 minus ๐.

Letโs begin by sketching the
straight line, noting it makes an angle of 135 degrees with the positive
๐ฅ-axis. We are told in the question that
the line passes through the point with coordinates one, negative 15. And we recall that the slope of the
line that makes an angle ๐ with the positive ๐ฅ-axis is given by tan ๐. As already mentioned, this angle ๐
is 135 degrees. And tan of 135 degrees is equal to
negative one. This means that the slope of our
line is negative one.

This slope is equal to the rise
over the run, which leads to a rise of negative one and a run of one. As the run is the change in the
๐ฅ-values and the rise is the change in the ๐ฆ-values, this gives us a direction
vector of one, negative one. The parametric form of the equation
of a line passing through a point with coordinates ๐ฅ sub zero, ๐ฆ sub zero and
parallel to any direction vector ๐, ๐ is ๐ฅ is equal to ๐ฅ sub zero plus ๐๐ and
๐ฆ equals ๐ฆ sub zero plus ๐๐. As we now have these two pieces of
information, we can substitute in our values.

Firstly, we have ๐ฅ is equal to one
plus ๐. And secondly, we have ๐ฆ is equal
to negative 15 minus ๐. The correct answer is therefore
option (A). The parametric equations of the
straight line that makes an angle of 135 degrees with the positive ๐ฅ-axis and
passes through the point one, negative 15 is ๐ฅ equals one plus ๐ and ๐ฆ equals
negative 15 minus ๐.

In our final question, we are given
the direction vector of the line by means of the slope.

A straight line passes through the
point one, six and has a slope of one-half. Which of the following pairs of
parametric equations represents this straight line? Is it (A) ๐ฅ is equal to ๐ก plus
one, ๐ฆ is equal to two ๐ก plus six? (B) ๐ฅ is equal to ๐ก plus two, ๐ฆ
is equal to six ๐ก plus one. (C) ๐ฅ is equal to four ๐ก plus
one, ๐ฆ is equal to two ๐ก plus six. (D) ๐ฅ is equal to six ๐ก plus one,
๐ฆ is equal to ๐ก plus two. Or (E) ๐ฅ is equal to four ๐ก plus
one, ๐ฆ is equal to ๐ก plus six.

We begin by recalling that the
parametric form of the equation of a line passing through a point with coordinates
๐ฅ sub zero, ๐ฆ sub zero and parallel to the direction vector with components ๐, ๐
is ๐ฅ is equal to ๐๐ก plus ๐ฅ sub zero and ๐ฆ is equal to ๐๐ก plus ๐ฆ sub
zero. We are told that the line passes
through the point one, six. Therefore, ๐ฅ sub zero equals one
and ๐ฆ sub zero equals six. We are also told it has a slope of
one-half. We can use this information to find
the direction vector of the line. Once we have found this, we can
substitute the values of ๐ and ๐ to establish the parametric equations for ๐ฅ and
๐ฆ.

Recalling that the slope is equal
to the rise over the run, as the slope of our line is one-half, the rise equals one
and the run equals two. This leads to a direction vector of
two, one. This can be demonstrated on the
๐ฅ๐ฆ-plane as shown. As the direction vector is equal to
two, one, the values of ๐ and ๐ in our parametric equations are two and one,
respectively.

Using the direction vector two, one
and the given point one, six, we can write the parametric form as ๐ฅ is equal to two
๐ก plus one and ๐ฆ is equal to ๐ก plus six. We note at this point that this
parametric form does not match any of the given options.

Letโs now clear some space and see
how we can overcome this. We need to choose an alternate
direction vector that is parallel to the vector two, one, for example, four, two;
six, three; eight, four; and so on. We can identify the direction
vector used for each option by recalling that the ๐ฅ- and ๐ฆ-components of the
direction vector can be obtained from the coefficients of ๐ก.

In option (A), the direction vector
is one, two. Option (B) has a direction vector
one, six. Neither of these are parallel to
the direction vector two, one. The direction vector of option (C)
is four, two. This is parallel to the direction
vector two, one, as we have multiplied both components by two. The direction vectors in options
(D) and (E) are six, one and four, one, respectively, neither of which is parallel
to two, one. Using the direction vector four,
two along with the point one, six, we have parametric equations ๐ฅ is equal to four
๐ก plus one and ๐ฆ is equal to two ๐ก plus six. The correct answer from the options
given is (C).

We will now summarize the key
points from this video. The parametric form of a straight
line gives ๐ฅ- and ๐ฆ-coordinates of each point on the line as a function of the
parameter. The parametric form of the equation
of a line passing through the point with coordinates ๐ฅ sub zero, ๐ฆ sub zero and
parallel to the direction vector ๐, which is equal to ๐, ๐, is ๐ฅ is equal to ๐ฅ
sub zero plus ๐๐ก and ๐ฆ is equal to ๐ฆ sub zero plus ๐๐ก. Any point on the line may be used
to obtain the parametric equations of the line. Also, the direction vector may be
replaced by any scalar multiple of the vector. This means that the parametric form
of the equation of a line will be nonunique.