### Video Transcript

Which of the following is a vector
equation for the line that passes through the midpoint of line segment π΄π΅, where
π΄ equals five, negative two and π΅ equals three, eight, and point three, one? Is it (A) vector π« equals five,
negative two plus π times seven, four? Is it (B) vector π« equals four,
three plus π times negative one, negative two? (C) Vector π« equals negative four,
negative three plus π times two, negative 10. (D) π« equals four, three plus π
times seven, four. Or is it (E) π« equals negative
one, negative two plus π four, three?

Letβs begin by recalling how we
represent the vector equation of a straight line. The vector equation of a straight
line can be represented in the form π« equals π« sub naught plus π‘ times π. π« sub naught is the position
vector of a point on the line. For this reason, π« sub naught can
take any number of values. We just need to make sure that the
point itself does lie on the line. π‘ is a scalar, a constant
scalar. And then π is the direction vector
of our line. So, our job is going to be to
identify the position vector and the direction vector. We might notice that in this
question, each scalar is given as the letter π instead of the letter π‘. And thatβs absolutely fine. Weβll use π throughout the
remainder of this question.

So, to find the position vector,
letβs identify a single point on the line. We could use the point three,
one. But weβre also told that our line
passes through the midpoint of line segment π΄π΅. We know that the position vector of
point three, one is three, one. And none of the lines in this
question have this as their position vector. So thatβs a good indication that
weβre going to need to find the midpoint of π΄π΅. To do so, we recall that we can
find the midpoint of a pair of coordinates by adding those coordinates and dividing
by two. In particular, the midpoint of π₯
sub one, π¦ sub one and π₯ sub two, π¦ sub two is π₯ sub one plus π₯ sub two over
two, π¦ sub one plus π¦ sub two over two.

Weβre looking to find the midpoint
of π΄ and π΅, so thatβs the midpoint of five, negative two and three, eight. Thatβs five plus three over two,
negative two plus eight over two. Five plus three is eight, and eight
divided by two is four. Similarly, negative two plus eight
divided by two is three. So, we found the midpoint of line
segment π΄π΅. Since the position vector of the
point on our line tells us how to get there from the origin, we know the position
vector π« sub naught can be given by the vector four, three.

Now that we have the position
vector, letβs identify the direction vector. We can find the direction vector by
finding the difference between the position vectors of two points on our line. We know that the line passes
through three, one and four, three. So, the direction vector could be
three, one minus four, three. Itβs worth noting that we can do
this in reverse. In other words, we could find four,
three minus three, one. When we multiply by the scalar,
that scalar can take any value. And so really, the direction vector
can be any multiple of this difference.

However, we might notice that if we
subtract the vector four, three from three, one, weβll get one of the direction
vectors in our question. In fact, subtracting the individual
components, we get the vector negative one, negative two. So taking the direction vector to
be negative one, negative two and the position vector of a point on our line to be
four, three and finally taking the scalar to be π, we get the equation of our line
to be four, three plus π times negative one, negative two. Thatβs equivalent to option
(B).