### Video Transcript

Consider the points π΄ equals
negative one, one and π΅ equals four, two. Parameterize the segment π΄π΅,
where π‘ is greater than or equal to zero and less than or equal to one.

Letβs begin by simply sketching the
diagram, showing our points and our line segment π΄π΅. It looks a little something like
this. Now, we know weβre going to go from
left to right generally. So weβll begin by letting π‘ be
equal to zero at the first point, at negative one, one. And that means we need to let π‘ be
equal to one at the other end of our interval at the point four, two. Now, weβre going to use the vector
form for the equation of a line. Itβs π equals π₯ nought, π¦ nought
plus π‘ times ππ.

Now, whilst this is an equation of
a line in two dimensions, this process can be extended to working with a line in
three dimensions. Weβll next say that the vector π
is of the form π₯, π¦. Then, weβll combine the vectors on
the right to π₯ nought plus π‘π, π¦ nought plus π‘π. And we see that the only way for
the vector on the left to be equal to the vector on the right is if its component
parts are equal. That is, if π₯ is equal to π₯
nought plus π‘π and π¦ is equal to π¦ naught plus π‘π. In fact, this pair of equations is
called the parametric Form of the equation of a line. Now, we can use these along with
the information in our question to parameterize the line segment from π to π.

Letβs go back to what we said
earlier. We said π‘ is equal to zero at
negative one, one. When π‘ is equal to zero, this
gives us the values for π₯ nought, π¦ nought. So π₯ nought must be equal to
negative one and π¦ nought must be equal to one. So we see the π₯ is equal to
negative one plus π‘π and π¦ is equal to one plus π‘π. Weβll now use the fact that when π‘
is equal to one, π₯ is equal to four and π¦ is equal to two. So our first equation becomes four
equals negative one plus one π. And our second equation becomes two
equals one plus one π. By adding one to both sides of our
first equation, we find π to be equal to five. And by subtracting one from both
sides of our second, we find π is equal to one. And we have our parametric
equations to describe the line segment π΄ to π΅ for values of π‘ between zero and
one. They are π₯ equals five π‘ minus
one and π¦ equals π‘ plus one.