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. Let’s begin by recalling the vector
form of a straight line. The vector form of a straight line
passing through the point 𝐴 and parallel to the direction vector 𝐝 is 𝐫 is equal
to 𝐎𝐀 plus 𝑡 multiplied by 𝐝. 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 𝐫. Each value of the parameter 𝑡
gives the position vector of one point on the line.
If we consider a line passing
through the point 𝐴 with coordinates 𝑥 sub zero, 𝑦 sub zero and parallel to the
direction vector 𝐝 with components 𝑎 and 𝑏, then the vector form of the equation
of the line is given by 𝐫 is equal to 𝑥 sub zero, 𝑦 sub zero plus 𝑡 multiplied
by 𝑎, 𝑏. Simplifying the right-hand side of
our equation, we get the components 𝑥 sub zero plus 𝑎𝑡 and 𝑦 sub zero plus
𝑏𝑡. We can then write the position
vector on the left-hand side in terms of its 𝑥- and 𝑦-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 𝐴 with coordinates 𝑥 sub zero, 𝑦 sub zero and
parallel to the direction vector 𝐝 is 𝑥 is equal to 𝑥 sub zero plus 𝑎𝑡, 𝑦 is
equal to 𝑦 sub zero plus 𝑏𝑡. In our first question, we will look
at an example of this in practice.
Straight line 𝐿 passes through the
point 𝑁 with coordinates three, four and has a direction vector 𝐮 equal to two,
negative five. Then the parametric equations of
line 𝐿 are what.
We begin by recalling that the
parametric form of the equation of a line passing through the point 𝑥 sub zero, 𝑦
sub zero and parallel to the direction vector 𝑎, 𝑏 is 𝑥 is equal to 𝑥 sub zero
plus 𝑎𝑡 and 𝑦 is equal to 𝑦 sub zero plus 𝑏𝑡. We are told in the question that
the straight line 𝐿 passes through the point with coordinates three, four. This means that our value of 𝑥 sub
zero is three and 𝑦 sub zero is equal to four. We are also given a direction
vector 𝐮 such that 𝑎 is equal to two and 𝑏 is negative five.
Substituting in our values of 𝑥
sub zero and 𝑎, we get 𝑥 is equal to three plus two 𝑡. And substituting the values of 𝑦
sub zero and 𝑏, we get 𝑦 is equal to four minus five 𝑡. It is important to note that we
could replace the letter 𝑡 with any other letter as the parameter. For example, 𝑥 is equal to three
plus two 𝑘 and 𝑦 is equal to four minus five 𝑘 is also a valid solution. We can therefore conclude that
these are the parametric equations of line 𝐿.
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 𝐫 of 𝑡 is equal to 𝑡 multiplied by five, two plus negative one,
three. Which of the following pairs of
parametric equations represents this straight line? Is it (A) 𝑥 is equal to five 𝑡
plus two, 𝑦 is equal to negative 𝑡 plus three? (B) 𝑥 is equal to five 𝑡 minus
one, 𝑦 is equal to two 𝑡 plus three. (C) 𝑥 is equal to three 𝑡 plus
two, 𝑦 is equal to negative 𝑡 plus five. (D) 𝑥 is equal to negative 𝑡 plus
five, 𝑦 is equal to three 𝑡 plus two. Or (E) 𝑥 is equal to two 𝑡 plus
three, 𝑦 is equal to five 𝑡 minus one.
We begin this question by recalling
that the vector form of the equation of a line is 𝐫 is equal to 𝑥 sub zero, 𝑦 sub
zero plus 𝑡 multiplied by 𝑎, 𝑏, where 𝑥 sub zero, 𝑦 sub zero is the position
vector of a point on the line and 𝑎, 𝑏 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 𝑡 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 𝑥 equals 𝑥 sub zero plus 𝑎𝑡 and 𝑦 equals 𝑦
sub zero plus 𝑏𝑡. Substituting in the values from
this question, we have 𝑥 is equal to negative one plus five 𝑡 and 𝑦 is equal to
three plus two 𝑡. Noting the way the equations have
been written in the five options, we have 𝑥 is equal to five 𝑡 minus one and 𝑦 is
equal to two 𝑡 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 𝑥 equals two and 𝑦 equals negative
two 𝑘 plus four is what.
We begin by recalling that the
parametric form of the equation of a line passing through the point with coordinates
𝑥 sub zero, 𝑦 sub zero and parallel to the direction vector with components 𝑎, 𝑏
is 𝑥 is equal to 𝑥 sub zero plus 𝑎𝑡 and 𝑦 is equal to 𝑦 sub zero plus
𝑏𝑡. We are given the parametric
equations 𝑥 equals two and 𝑦 equals negative two 𝑘 plus four.
Noting that the parameter here is
𝑘, we can rewrite the general equations as shown. Comparing terms, we see that 𝑥 sub
zero is equal to two. 𝑎 is equal to zero, as there is no
𝑘 term in our first parametric equation. We also have 𝑦 sub zero is equal
to four and 𝑏 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 𝑥 equals two
and 𝑦 equals negative two 𝑘 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 𝐫 is equal to
two, four plus 𝑘 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 𝑥-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 𝑥-axis and
passes through the point one, negative 15. Is it (A) 𝑥 is equal to one plus
𝑘, 𝑦 is equal to negative 15 minus 𝑘? (B) 𝑥 is equal to one plus 𝑘, 𝑦
is equal to one minus 15𝑘. Is it (C) 𝑥 is equal to negative
15 minus 𝑘, 𝑦 is equal to one plus 𝑘? Or (D) 𝑥 is equal to one, 𝑦 is
equal to negative 15 minus 𝑘.
Let’s begin by sketching the
straight line, noting it makes an angle of 135 degrees with the positive
𝑥-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 𝜃 with the positive 𝑥-axis is given by tan 𝜃. As already mentioned, this angle 𝜃
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
𝑥-values and the rise is the change in the 𝑦-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 𝑥 sub zero, 𝑦 sub zero and
parallel to any direction vector 𝑎, 𝑏 is 𝑥 is equal to 𝑥 sub zero plus 𝑎𝑘 and
𝑦 equals 𝑦 sub zero plus 𝑏𝑘. As we now have these two pieces of
information, we can substitute in our values.
Firstly, we have 𝑥 is equal to one
plus 𝑘. And secondly, we have 𝑦 is equal
to negative 15 minus 𝑘. The correct answer is therefore
option (A). The parametric equations of the
straight line that makes an angle of 135 degrees with the positive 𝑥-axis and
passes through the point one, negative 15 is 𝑥 equals one plus 𝑘 and 𝑦 equals
negative 15 minus 𝑘.
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) 𝑥 is equal to 𝑡 plus
one, 𝑦 is equal to two 𝑡 plus six? (B) 𝑥 is equal to 𝑡 plus two, 𝑦
is equal to six 𝑡 plus one. (C) 𝑥 is equal to four 𝑡 plus
one, 𝑦 is equal to two 𝑡 plus six. (D) 𝑥 is equal to six 𝑡 plus one,
𝑦 is equal to 𝑡 plus two. Or (E) 𝑥 is equal to four 𝑡 plus
one, 𝑦 is equal to 𝑡 plus six.
We begin by recalling that the
parametric form of the equation of a line passing through a point with coordinates
𝑥 sub zero, 𝑦 sub zero and parallel to the direction vector with components 𝑎, 𝑏
is 𝑥 is equal to 𝑎𝑡 plus 𝑥 sub zero and 𝑦 is equal to 𝑏𝑡 plus 𝑦 sub
zero. We are told that the line passes
through the point one, six. Therefore, 𝑥 sub zero equals one
and 𝑦 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 𝑎 and 𝑏 to establish the parametric equations for 𝑥 and
𝑦.
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
𝑥𝑦-plane as shown. As the direction vector is equal to
two, one, the values of 𝑎 and 𝑏 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 𝑥 is equal to two
𝑡 plus one and 𝑦 is equal to 𝑡 plus six. We note at this point that this
parametric form does not match any of the given options.
Let’s 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 𝑥- and 𝑦-components of the
direction vector can be obtained from the coefficients of 𝑡.
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 𝑥 is equal to four
𝑡 plus one and 𝑦 is equal to two 𝑡 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 𝑥- and 𝑦-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 𝑥 sub zero, 𝑦 sub zero and
parallel to the direction vector 𝐝, which is equal to 𝑎, 𝑏, is 𝑥 is equal to 𝑥
sub zero plus 𝑎𝑡 and 𝑦 is equal to 𝑦 sub zero plus 𝑏𝑡. 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.
