### Video Transcript

In this video, we will learn how to
find the coordinates of a point that divides a line segment on the coordinate plane
in a given ratio using the section formula. Letβs begin by reviewing some
important terminology, firstly, line segments.

A line segment is a part of a line
thatβs bounded by two distinct endpoints. For example, here we have the line
segment π΄π΅. We can even consider a line segment
on the coordinate plane. If we have the coordinates of π΄
and π΅, we can even find the midpoint of this line segment by using the midpoint
formula. In this video, however, we want to
go one step further than simply dividing a line segment into two pieces. And weβll see how we can find the
point which divides a line segment into a given ratio.

Letβs see what that would look
like. Here we have a line segment π΄π΅,
and weβre given the coordinates of these two points. Letβs then say that we need to find
the coordinates of this point π which divides line segment π΄π΅ in the ratio π to
π. Is there a way in which we can find
the coordinates of point π? Well, to do this, weβll need to
start by constructing two right triangles. In these right triangles, one of
them will have a hypotenuse of π and the other one will have a hypotenuse of
π. Note that these two triangles will
be similar since the corresponding angles of the triangle are congruent. In similar triangles, the ratio of
corresponding sides will be equal.

To find the distance of π from π΄,
it will be π over π plus π multiplied by the length of π΄π΅. We can use this to find the π₯- and
π¦-values of point π. So letβs begin with how we would
find the π₯-value. π₯ is equal to π₯ sub one, thatβs
the π₯-value in our π΄-point, plus π over π plus π multiplied by π₯ sub two
subtract π₯ sub one. This second part of this equation
represents our distance of π from point π΄ in terms of the π₯-values. In order to simplify this, we now
need to make sure that our value of π₯ one is also as a fraction over π plus
π. If we multiply π₯ sub one by π
plus π on the numerator and denominator, weβll get π₯ is equal to π plus π times
π₯ sub one over π plus π plus π multiplied by π₯ sub two subtract π₯ sub one over
π plus π.

We can then expand both sets of
parentheses, and we notice that we have ππ₯ sub one subtract ππ₯ sub one. Weβre then left with π₯ is equal to
ππ₯ sub two plus ππ₯ sub one over π plus π. We could then perform exactly the
same process to find the value of π¦. This time, weβd swap our π₯ sub one
and π₯ sub two values for π¦ sub one and π¦ sub two, respectively. We would then find that π¦ is equal
to ππ¦ sub two plus ππ¦ sub one over π plus π. And so weβve now found the π₯-value
and the π¦-value of point π, which divides our line segment. Letβs write this in a nicer way so
that we can make a note of it.

If π΄ with coordinates π₯ sub one,
π¦ sub one and π΅ with coordinates π₯ sub two, π¦ sub two and point π divides line
segment π΄π΅ such that π΄π to ππ΅ equals the ratio π to π, then π has
coordinates. π equals ππ₯ sub two plus ππ₯
sub one over π plus π, ππ¦ sub two plus ππ¦ sub one over π plus π. We can now see how our π₯- and
π¦-values come together to create this coordinate of π. This formula does look quite
difficult, but when we put it into practice, itβs really not so difficult. Itβs not the easiest formula to
remember, but as we go through this video, weβll write it out in each question. And hopefully, by the end of it,
weβll have learned it a little bit more. Letβs take a look at our first
question.

If the coordinates of π΄ and π΅ are
five, five and negative one, negative four, respectively, find the coordinates of
the point πΆ that divides vector π΄π΅ internally by the ratio two to one.

It might be sensible to begin a
question like this by plotting our two coordinates. We could either create a sketch of
this graph or use grid paper. So here we have a grid drawn. The point π΄ is at the coordinate
five, five and π΅ is at negative one, negative four. As weβre told that thereβs a vector
π΄π΅, we can join our two points. Weβre then asked to find the
coordinates of a point πΆ which divides our vector π΄π΅ internally in the ratio two
to one. This means that weβll need to use
the section formula. For point π΄ with coordinates π₯
sub one, π¦ sub one and point π΅ with coordinates π₯ sub two, π¦ sub two, the point
π which divides line segment π΄π΅ in the ratio π to π has coordinates π is equal
to π times π₯ sub two plus π times π₯ sub one over π plus π, π times π¦ sub two
plus π times π¦ sub one over π plus π.

The key pieces of information that
we need for the section formula are the two coordinates of π΄ and π΅ and the
ratio. As the direction is important,
weβre going from π΄ to π΅, then our π₯ sub one, π¦ sub one values must go with point
π΄. The ordering of our ratio values of
π and π is also important, so π will be two and π will be one. We can then plug these values into
the formula and simplify. For the π₯-value of point π, we
have two multiplied by negative one which is negative two plus five gives us three,
and two plus one on the denominator of course becomes three. For the π¦-value, we have two
multiplied by negative four, which is negative eight, plus one times five, which is
five, gives us a value of negative three on the numerator. And our denominator will also be
three. Simplifying these two fractions
then, three over three is one and negative three over three is negative one.

So now we know that point πΆ which
divides this vector π΄π΅ in the ratio two to one is the coordinate one, negative
one. A really good check of the answer
if we have drawn a graph on grid paper is to check that the coordinate actually does
lie on the line, and it does. Here we can see point πΆ with
coordinates one, negative one. We can also see that line segment
π΄π΅ is divided in the ratio two to one. Weβve therefore verified the
coordinate one, negative one is the answer.

Letβs take a look at another
question.

The coordinates of π΄ and π΅ and
are one, nine and nine, nine, respectively. Determine the coordinates of the
points that divide line segment π΄π΅ into four equal parts.

Letβs start this question by
imagining this line segment joining π΄ and π΅. We can consider this line split
into four parts. And if we wanted to find this first
point, we could consider how we would split this line segment π΄π΅ in a ratio one to
three. The second point could be found by
splitting it in the ratio two to two or even by finding the midpoint. We could find the coordinates of
the third point by splitting the line segment π΄π΅ in the ratio three to one. We could actually do these three
pieces of working using the section formula. This tells us that for any two
points π΄ and π΅ with coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two,
respectively, the point π which divides line segment π΄π΅ in the ratio π to π has
coordinates π equal to ππ₯ sub two plus ππ₯ sub one over π plus π, ππ¦ sub two
plus ππ¦ sub one over π plus π.

In this question, we need to apply
the section formula three times to find these three different coordinates with their
different ratios. There is actually an easy way if we
consider these two coordinates one, nine and nine, nine. We can plot these as shown and even
create the line segment π΄π΅. As this is a horizontal line, itβs
a little bit easier to work out the distance of π΄π΅. It would be eight units long. We can then divide this length into
four equal parts, and so weβll get the coordinates three, nine; five, nine; and
seven, nine. This will be our answer for the
coordinates that divide line segment π΄π΅ into four equal parts. We could have done this in the same
way using the section formula, but it would have taken a lot longer.

Letβs take a look at another
question.

If πΆ is an element of π΄π΅ and
vector π΄π΅ equals three times vector πΆπ΅, then πΆ divides vector π΅π΄ by the ratio
blank. Option (A) two to one, option (B)
one to two, option (C) one to three, option (D) three to one.

Thereβs quite a lot of information
in this question. But letβs start with the fact that
we have this line segment π΄π΅, which we could model like this. Weβre told that πΆ is an element of
π΄π΅, so that means that there will be a point πΆ somewhere on this line. If vector π΄π΅ is equal to three
times vector πΆπ΅, then that means that three of this length πΆπ΅ would make up the
length of π΄π΅. We could divide our length π΄π΅
into three pieces, but the question is, is πΆ here or is πΆ here? If we consider if πΆ is at this
lower point, then the length πΆπ΅ would look like this. But if we were to multiply πΆπ΅ by
three, we wouldnβt get the length of π΄π΅. We can then say that πΆ must be
here, closer to π΅, as this length of πΆπ΅ would fit. Three lots of πΆπ΅ would give us
π΄π΅.

We now need to work out the
question of how πΆ divides this vector π΅π΄. We can then say if π΅πΆ is one unit
length long, then π΄πΆ would be equivalent to two of these lengths. So do we write this ratio as two to
one or one to two? Well, the direction here is very
important. Weβre given the vector π΅π΄, so
that means that weβre going from π΅ to π΄. We can therefore give our answer
that itβs the ratio one to two, which is given in option (B). Note that if we had been given the
vector π΄π΅ instead, then that wouldβve been the ratio two to one. But here, since πΆ is dividing
vector π΅π΄, then itβs the ratio one to two.

Weβll now look at one final
question.

A bus is traveling from city π΄ at
coordinates 10, negative 10 to city π΅ at coordinates negative eight, eight. Its first stop is at πΆ, which is
halfway between the cities. Its second stop is at π·, which is
two-thirds of the way from π΄ to π΅. What are the coordinates of πΆ and
π·?

In this problem question, we have a
bus which is traveling from π΄ to π΅. It stops halfway, which is at the
point πΆ, and then it stops again at point π·, which is two-thirds of the way from
π΄ to π΅. We might want to start this by
modeling our coordinates on a graph. So here we have π΄ at 10, negative
10 and π΅ at negative eight, eight. We could even join these with a
line. The halfway point on this bus
journey is at point πΆ. If weβd used a graph paper for this
and we had a nice integer result, we might be able to read the coordinate of point
πΆ directly from the graph. But letβs see if we can solve this
using the formula to find the midpoint of a line segment. This formula tells us for the two
points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two, the midpoint of the line
joining these can be found at the coordinate π₯ sub one plus π₯ sub two over two, π¦
sub one plus π¦ sub two over two.

We can designate the point π΄ with
the π₯ sub one, π¦ sub one values and point π΅ with the π₯ sub two and π¦ sub two
values, and we then plug these into the formula. This gives us that the π₯-value of
our midpoint will be 10 plus negative eight over two and the π¦-value will be
negative 10 plus eight over two. The midpoint can then be written as
two over two, negative two over two, which of course is one, negative one. We have therefore found our first
answer. The coordinate of πΆ, which is the
midpoint, is at one, negative one.

Letβs clear some space and see if
we can find the coordinates of π·. Weβre given that π· is two-thirds
of the way from π΄ to π΅. We could split our line into three
equal parts, and we know that π· is two of these three parts along. Although we have a midpoint formula
to find a value halfway along a line, we donβt have a formula to find a value
two-thirds of the way along the line. We might remember, however, that
thereβs the section formula, which allows us to split a line segment into a given
ratio. This tells us that for two points
π΄ with coordinates π₯ sub one, π¦ sub one and π΅ with coordinates π₯ sub two, π¦
sub two, the point π, which divides line segment π΄π΅ in the ratio π to π, has
coordinates π equal to ππ₯ sub two plus π π₯ sub one over π plus π, ππ¦ sub
two plus ππ¦ sub one over π plus π.

In order to find the coordinates of
our point π· which divides this line segment π΄π΅, we need to know the ratio. Since we were told that π· is
two-thirds of the way from π΄ to π΅, we split π΄π΅ into three parts. From π΄ to π·, itβs two out of
those three parts, and from π· to π΅, itβs the remaining one part. Point π· therefore splits the line
segment π΄ to π΅ in the ratio two to one. To use this section formula, we
need the letters π to π which is the ratio two to one, and we need our two
coordinates π΄ and π΅.

We can therefore plug in our values
into the formula to give us that the π₯-value of coordinate π· is two times negative
eight plus one times 10 over two plus one. And the π¦-value of this coordinate
is two times eight plus one times negative 10 over two plus one. Simplifying these values gives us
that the coordinates of π· are negative six three, six three, which simplifies to
negative two, two. We can therefore give our two
answers πΆ is at the coordinates one, negative one and π· is at the coordinates
negative two, two.

We can now summarize what weβve
learned in this video. Firstly, we saw the section
formula, which tells us that for point π΄ with coordinates π₯ sub one, π¦ sub one
and point π΅ with coordinates π₯ sub two, π¦ sub two, the point π which divides
line segment π΄π΅ in the ratio π to π has coordinates as follows. The π₯-value is π times π₯ sub two
plus π times π₯ sub one over π plus π, and the π¦-value is π times π¦ sub two
plus π times π¦ sub one over π plus π.

We also saw that when weβre
partitioning or dividing a line into two equal pieces, then we can use the formula
to find the midpoint of a line segment. The midpoint of the line joining
the two coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two has the
coordinates π₯ sub one plus π₯ sub two over two, π¦ sub one plus π¦ sub two over
two. Finally, a handy tip: itβs very
useful to draw the graphs and plot any points that weβre given. This will help us think logically
through the problems and is useful when weβre finding the ratios in a line
segment.