### Video Transcript

In this video, we will learn how to
find the distance between two points on the coordinate plane using the Pythagorean
theorem.

Letβs begin by recalling the
Pythagorean theorem. And then weβll see how we can use
this theorem to create the formula that allows us to find the distance between any
two coordinates. The Pythagorean theorem states that
in any right triangle, the square of the hypotenuse is equal to the sum of the
squares of the two shorter sides. We often see this given
diagrammatically with the two shorter sides of the right triangle labeled with π
and π and the hypotenuse, thatβs the longest side, labeled with the letter π. So, by the Pythagorean theorem, we
can say that π squared plus π squared is equal to π squared.

Now, letβs see how we can use this
theorem to find the distance between two points on a coordinate grid. Letβs take this example of the line
joining the points negative two, one and three, four. Notice that we can create a right
triangle with the hypotenuse as the line joining the two points. A right angle is created at the
coordinates three, one, where the vertical line from three, four and the horizontal
line from negative two, one meet. We can determine that the
horizontal distance from negative two, one to three, one is five units and the
vertical distance from three, one to three, four is three units.

We now have enough information to
apply the Pythagorean theorem. If we take π to be three units and
π to be five units, then by using the Pythagorean theorem, we have three squared
plus five squared is equal to π squared. By simplifying this, we have that
34 is equal to π squared. Taking the square root of both
sides, we have that the square root of 34 is equal to π. That means that we have calculated
that the distance between these two coordinates is the square root of 34 length
units.

So, at a very basic level, we can
see how counting the squares on vertical and horizontal distances can allow us to
calculate the distance between two coordinates using the Pythagorean theorem. But of course this is not a very
practical solution for finding the distance between any two coordinates. We need to see how we can use the
Pythagorean theorem without drawing a graph.

Letβs consider a more general
situation where we want to find the distance π between two coordinates called π₯
sub one, π¦ sub one and π₯ sub two, π¦ sub two. As weβve seen previously, we know
that we can create a right triangle using horizontal and vertical lines. The third vertex of this triangle
would be at the coordinates π₯ sub two, π¦ sub one.

Now, we need to work out the length
of these two line segments which form the two shorter sides. The horizontal distance between the
points is π₯ sub two minus π₯ sub one, and the vertical distance is π¦ sub two minus
π¦ sub one. However, the value of these
distances must always be positive in order for our method to work. Therefore, in order to generalize
for any positive or negative values of π₯ sub one, π¦ sub one, π₯ sub two, or π¦ sub
two, we need to use absolute value symbols. This will indicate that the length
is a positive number because the absolute value of any number is positive.

Then, when we apply the Pythagorean
theorem, we can take either of these sides to be π or π, so taking π to be the
absolute value of π₯ sub two minus π₯ sub one and π to be the absolute value of π¦
sub two minus π¦ sub one. We can then substitute these values
into the Pythagorean theorem to find the distance π between the points. But if youβve seen this formula
before, you might have noticed that there are no absolute value symbols. Thatβs because our terms of π₯ sub
two minus π₯ sub one and π¦ sub two minus π¦ sub one are being squared. Any square number will be positive,
and hence we will not have any issues with any negative distances.

When we fill these into the
Pythagorean theorem then, we can simply write that π₯ sub two minus π₯ sub one
squared plus π¦ sub two minus π¦ sub one squared is equal to π squared. And as we donβt want π squared β
we want π β we need to take the square root of both sides of this equation. This means that we have now derived
a formula to help us calculate the distance between any two points on the coordinate
plane.

Letβs make a note of this formula,
which is often referred to as the distance formula. The distance π between two points
with coordinates π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is given by π
equals the square root of π₯ sub two minus π₯ sub one squared plus π¦ sub two minus
π¦ sub one squared.

Now, letβs see how we can apply
this formula in the following examples.

Find the distance between the point
negative two, four and the point of origin.

We can recall the distance formula
that the distance π between π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is
given by π equals the square root of π₯ sub two minus π₯ sub one squared plus π¦
sub two minus π¦ sub one squared. We are given two points. One is negative two, four, and the
other is the point of origin, which can be given as the coordinates zero, zero. Letβs define zero, zero with the π₯
sub one, π¦ sub one values and negative two, four with the π₯ sub two, π¦ sub two
values. But it doesnβt matter which
coordinate we define with these values.

Substituting these values into the
formula gives us that π is equal to the square root of negative two minus zero
squared plus four minus zero squared. We can then simplify each set of
parentheses and square the values, giving us that the distance π is root 20 length
units. We can leave our answer in this
radical form as we werenβt required to give our answer as a decimal. But we can of course simplify this
radical further. As root 20 can be written as root
four times root five, then this can be simplified to give us the answer for the
distance as two root five length units.

Before we finish with this
question, itβs worth demonstrating that it doesnβt matter which coordinates we
designate with the π₯ sub one, π¦ sub one values and which we designate with the π₯
sub two, π¦ sub two values. Letβs see what happens if we switch
these values. We can define the point negative
two, four to have the π₯ sub one, π¦ sub one values. Filling these values into the
formula would give us that π is equal to the square root of zero minus negative two
squared plus zero minus four squared. When we simplify this, we can see
how we end up with the same answer of two root five length units. This is as a result of squaring
each of the values of π¦ sub two minus π¦ sub one and π₯ sub two minus π₯ sub one,
which always produces a positive result. Using either calculation gives us
the solution that the distance between negative two, four and the point of origin is
two root five length units.

In the next example, weβll see how
to find a missing coordinate given the distance between two points.

The distance between π, five and
one, one is five. What are the possible values of
π?

In order to answer this question,
weβll need to recall and use the distance formula. This tells us that the distance π
between π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two is given by π equals the
square root of π₯ sub two minus π₯ sub one squared plus π¦ sub two minus π¦ sub one
squared. Very commonly, we simply use this
formula to work out the distance between two points. However, in this question, weβre
told that the distance is five, so that means that π is equal to five.

We can then define the coordinates
one, one to have the π₯ sub one, π¦ sub one values, although it wouldnβt matter if
we define these with π₯ sub two, π¦ sub two. The second coordinate, which has
this unknown value of π, can be given the π₯ sub two, π¦ sub two values. Substituting these in to the
formula gives five is equal to the square root of π minus one squared plus five
minus one squared. Simplifying within the square root,
we know that five minus one is four and four squared is 16.

As we want to work out the value of
π, the next stage is to square both sides. This gives us five squared is equal
to π minus one squared plus 16. We know that five squared is equal
to 25. And so in the next stage of
simplification, we subtract 16 from both sides of the equation, leaving us with nine
is equal to π minus one squared. Next, we take the square root of
both sides of the equation. But since the square of a negative
number is still positive, we must account for both the positive and negative
roots. And we can do this using the plus
minus sign.

By finding the square root of nine,
weβre left with the equation plus or minus three is equal to π minus one. This means that weβll have two
equations. π minus one is equal to positive
three, and π minus one is equal to negative three. In the case when π minus one
equals positive three, we can solve for π to establish that π is equal to
four. And in the case when π minus one
is equal to negative three, then π is equal to negative two.

We can therefore give the answer
that the possible values of π are π equals negative two or four. Perhaps the best way to understand
why there is more than one solution is by drawing a quick sketch. We were given in the question that
there is a complete coordinate one, one. And we were also given the
coordinate π, five. We can think of π, five as a point
with an unknown π₯-coordinate but a π¦-coordinate of five. That means that the missing
coordinate must lie somewhere on the line with the equation π¦ is equal to five.

We were given that the distance of
one, one from π, five is five units. So there are in fact two
possibilities. The coordinates of the two points
at a distance of five units from one, one that lie on the line π¦ equals five are
negative two, five and four, five. Itβs worth noting that there are an
infinite number of possible coordinates that just lie a distance of five units from
one, one. All of these coordinates would
create a circle of radius five units from one, one. However, in this question, we were
constrained by the fact that the π¦-coordinate of π, five is five. So there were just two possible
solutions. These occur when π is equal to
negative two or four.

In the next example, weβll see a
problem involving a circle. But given that the circle is
plotted on the coordinate plane and given the center and a point on the
circumference, this means that we can apply the distance formula to calculate its
radius.

Point negative six, seven is on the
circle with center negative seven, negative one. Decide whether point negative
eight, negative nine is on, inside, or outside the circle.

Letβs begin this problem by
visualizing the information that weβre given. We are told that the point negative
six, seven is on the circle with center negative seven, negative one, which means
that negative six, seven lies on the circumference of this circle. We then need to determine where
this third point negative eight, negative nine lies in relation to the circle.

There are in fact three
possibilities. The first possibility is that
negative eight, negative nine is also on the circle. The second possibility is that
negative eight, negative nine lies inside the circle. Or thirdly, negative eight,
negative nine could lie outside the circle. Although we have drawn some
sketches here, theyβre not accurate enough to allow us to determine this. And even if we drew a very accurate
diagram, that would not be a good enough determination either. Instead, we need to find an
algebraic method to determine for sure where negative eight, negative nine lies. So, how do we compare where
negative eight, negative nine lies in relation to the circle?

The first thing we should realize
is that if we knew the distance from negative six, seven to the center negative
seven, negative one, that would represent the radius of the circle. Letβs say that this value is
π. If we then determine the distance
from the center to the point negative eight, negative nine, then we could compare
this to the length of the radius. If the distance is the same as the
radius, then negative eight, negative nine lies on the circle. Or if the distance is smaller than
the radius, then negative eight, negative nine is inside the circle. The final possibility is that the
distance is larger than the radius, which would make negative eight, negative nine
outside the circle.

We recall that we can find the
distance between two points π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two by
using the distance formula, which tells us that the distance π between two points
is given as the square root of π₯ sub two minus π₯ sub one squared plus π¦ sub two
minus π¦ sub one squared. Letβs then start by working out the
value of the length of this radius between negative seven, negative one and negative
six, seven. When we substitute these in to the
formula, we need to be very careful as there are lots of negative values of
coordinates here, so we must remember to include them all. We can then simplify this to give
us π is equal to one squared plus eight squared. We have then determined that π,
which represents the radius, is root 65 length units.

Now, letβs use the same formula to
work out the distance between the center negative seven, negative one and the point
negative eight, negative nine. This distance π can be found by
substituting the values in to the formula to give us the square root of negative
seven minus negative eight squared plus negative one minus negative nine
squared. When we simplify, we get a value of
root 65 length units for the distance between these two points. In fact, this length is exactly the
same as the length of the radius. We can therefore give the answer
that the point negative eight, negative nine must lie on the circle.

We can now summarize the key points
of this video. We began by recalling the
Pythagorean theorem, which tells us that in any right triangle, the square of the
hypotenuse is equal to the sum of the squares of the two shorter sides. We then derived the distance
formula, which gives us that the distance π between two points with coordinates π₯
sub one, π¦ sub one and π₯ sub two, π¦ sub two is given by π equals the square root
of π₯ sub two minus π₯ sub one squared plus π¦ sub two minus π¦ sub one squared. We note that when using the
distance formula with two coordinates, we can substitute either set of coordinates
for the π₯ sub one, π¦ sub one and π₯ sub two, π¦ sub two values. And finally, as we saw in the
second example, when using the distance formula to find an unknown coordinate at a
given distance from another set of coordinates, there may be more than one
solution.