Video Transcript
Find the parametric equation of the
line that passes through the midpoint of the line segment 𝐴𝐵, where 𝐴 is the
point two, negative one and 𝐵 is the point four, three, and point two, negative
three. Is it option (A) 𝑥 is equal to one
minus four 𝑘 and 𝑦 is equal to three minus 𝑘? Option (B) 𝑥 is equal to three
plus 𝑘 and 𝑦 is equal to one minus four 𝑘. Option (C) 𝑥 is equal to three
minus 𝑘 and 𝑦 is equal to one minus four 𝑘. Option (D) 𝑥 is equal to three
minus 𝑘 and 𝑦 is equal to one plus four 𝑘. Or is it option (E) 𝑥 is equal to
negative three minus 𝑘 and 𝑦 is equal to negative one minus four 𝑘?
In this question, we’re asked to
find the parametric equation of a line. So let’s start by recalling what we
mean by the parametric equation of a line. These are two equations of the form
𝑥 is equal to 𝑥 sub zero plus 𝑎 times 𝑘 and 𝑦 is equal to 𝑦 sub zero plus 𝑏
multiplied by 𝑘, where the point 𝑥 sub zero, 𝑦 sub zero is any point which lies
on the line and the vector 𝐚𝐛 is any nonzero vector which is parallel to the
line. This means we can actually find
infinitely many parametric equations which represent the same straight line, since
we can choose any point 𝑥 sub zero, 𝑦 sub zero which lies on the line and we can
choose any nonzero vector 𝐚𝐛 which is parallel to the line.
We can start by noting we’re told
that the line passes through the point with coordinates two, negative three. So we could set 𝑥 sub zero equal
to two and 𝑦 sub zero equal to negative three. However, we can notice something
interesting if we look at our options. In all five of the given options,
we can see that this is not the point given as 𝑥 sub zero, 𝑦 sub zero. So, although we can choose this for
our point 𝑥 sub zero, 𝑦 sub zero, it will not be one of the five given
options. So let’s not use this point.
Instead, we can also note that the
line is passing through the midpoint of the line segment 𝐴𝐵. So let’s find the coordinates of
the midpoint of line segment 𝐴𝐵. We’ll call this point 𝐶. We can find the coordinates of a
midpoint of a line segment by using the coordinates of its endpoints by just finding
the average of the 𝑥-coordinates and 𝑦-coordinates. The average of the 𝑥-coordinates
of points 𝐴 and 𝐵 is two plus four over two. And the average of the
𝑦-coordinates of points 𝐴 and 𝐵 is negative one plus three over two. And we can evaluate both of
these. We get that 𝐶 is the point three,
one.
And remember our line passes
through the point with coordinates three, one. So we can set 𝑥 sub zero equal to
three and 𝑦 sub zero equal to one in our parametric equations. But we’re not done yet. We also need to find the vector
parallel to the line. And to do this, we need to note
that our line passes through the point 𝐶 and it also passes through the point with
coordinates two, negative three. If we add this point onto our
diagram and we also add the line segment between point 𝐶 and the point two,
negative three, we get the following. And since we have the endpoints of
this line segment, we can use this to determine a vector parallel to this line.
We can just take the difference in
the position vectors of the two endpoints. However, the order we choose these
will determine the direction our vector is facing. For example, we can choose our
vector to terminate at the point two, negative three. For example, if we call this vector
𝐝, then we can find the components of vector 𝐝 by subtracting the position vector
of the initial point of this vector from the position vector of its terminal
point. 𝐝 is the vector two, negative
three minus the vector three, one. And we evaluate vector subtraction
component-wise. 𝐝 is the vector two minus three,
negative three minus one, which we can evaluate is the vector negative one, negative
four.
And before we use these values as
𝑎 and 𝑏, it’s worth noting we can choose any nonzero scalar multiple of this
vector, since this will still be parallel to the line. For example, we may want to choose
the vector one, four, multiplying this vector by negative one, which we can also
note is the exact result we would’ve got by choosing our points in the opposite
order, in other words, switching the direction of vector 𝐝.
However, none of this is
necessary. If we look at option (C), we can
see that the point which lies on the line is the point three, one, which we found is
point 𝐶. And the direction vector of the
line, which we find by looking at the coefficients of our parameters, is negative
one, negative four. So, if we set 𝑥 sub zero equal to
three, 𝑦 sub zero equal to one, 𝑎 equal to negative one, and 𝑏 equal to negative
four, we get the parametric equations 𝑥 is equal to three minus 𝑘 and 𝑦 is equal
to one minus four 𝑘, which is option (C).
