### Video Transcript

Consider the point π΄ with
rectangular coordinates negative four, seven. Calculate the distance π of
this point from the origin. Give your answer in exact
form. Find the angle π that the line
segment ππ΄ makes with the positive π₯-axis, giving your answer in radians to
two decimal places. Given that the point π΄ can be
expressed in polar form as π, π, which of the following is also a legitimate
polar form for point π΄? Option (a) π, π minus two π;
option (b) π, π minus π; option (c) negative π, π minus two π; option (d)
π, π plus three π; or option (e) negative π, π plus two π.

Weβre told that the point π΄
has rectangular coordinates negative four, seven. In fact, weβre shown this on a
diagram. We need to calculate the
distance this point π΄ is from the origin. Weβll call this distance
π. And we need to give this in an
exact form. In other words, π is just the
length of our line segment ππ΄. Thereβs several different
methods we could use to find this value. For example, we could use our
formula to calculate the distance between two points. However, since weβre given
these points in the diagram, we could also do this by using the Pythagorean
theorem.

Remember, the point π΄ is at
the coordinates negative four, seven. So if we drop a vertical line
from π΄ to the π₯-axis, we get the following right-angled triangle. We can see the height of this
right-angled triangle will be the π¦-coordinate of π΄ which is seven. And to find the width of this
right-angled triangle, we can see we go from π₯ is equal to zero to π₯ is equal
to negative four. So this right-angled triangle
has width four. So now, we have a right-angled
triangle with height seven and base four. And we need to find the length
of the hypotenuse π.

We can, of course, do this by
using the Pythagorean theorem. The Pythagorean theorem tells
us that π squared will be equal to four squared plus seven squared. And we can calculate four
squared plus seven squared to give us 65. And then we can solve for the
value of π by taking the square root of both sides of this equation. Remember, π represents the
length, so it wonβt be negative. So we take the positive square
root. So by taking the positive
square root of both sides of this equation, we get that π is equal to the
square root of 65. And remember, the question
wants us to give this length in an exact form. So we can just leave our answer
like this.

The next part of this question
wants us to find the angle π. And weβre told that this angle
is the angle that the line segment ππ΄ makes with the positive π₯-axis. And we need to give our answer
in radians to two decimal places. Once again, we can see that
this angle of π is given to us in the diagram. Thereβs a lot of different ways
of calculating the value of π. In this video, weβre only going
to go through one of them.

Consider the following
angle. We can see that this angle that
we just marked and π add together to give us half of a turn. And of course, we know a half
turn is π radians. So these two angles add
together to give us π. And if these two angles add
together to give us π, then this angle must be equal to π minus π. So instead of finding the angle
of π directly, weβll instead find the value of this angle, π minus π. To do this, we notice this is
an angle in a right-angled triangle, and we know the length of the opposite and
adjacent side of this angle. So, in fact, we can just find
this angle by using trigonometry.

We need to recall the tangent
of an angle in a right-angled triangle will be equal to the length of its
opposite side divided by the length of its adjacent side. So by using our right-angled
triangle with the angle π minus π, we get the tan of π minus π is equal to
seven divided by four. And remember, weβre trying to
find the value of π. So we need to take the inverse
tangent of both sides of this equation. Doing this, we get that π
minus π is equal to the inverse tan of seven divided by four.

Finally, all we need to do is
rearrange this equation to make π the subject. Weβll add π to both sides of
this equation and then subtract the inverse tangent of seven over four from both
sides of the equation. Doing this, we get that π is
equal to π minus the inverse tan of seven over four. And remember, we need to set
our calculator into radians mode. Doing this to two decimal
places, we get that π minus the inverse tan of seven over four is approximately
equal to 2.09. Therefore, we were able to show
to two decimal places the angle the line segment ππ΄ makes with the positive
π₯-axis π is equal to 2.09.

And now, weβre given some
information about the point π΄. Weβre told that the point π΄
can be expressed in polar form as π, π. And we just calculated the
values of π and π. And of course, we know that
polar forms for a point are not unique. We need to determine which of
the following is also a legitimate polar form of our point π΄. The first thing we notice is
options (a), (b), and (d) all contain a positive π, whereas options (c) and (e)
contain negative π. So letβs start by recalling
what makes two polar coordinates equal.

One way of having two sets of
polar coordinates represent the same point is to have the same value of π. However, they have two
different values of π which represent the same angle. And we know how to do this. For example, instead of using
the angle π, we can instead take one full turn around our π₯-axis and then
another turn at π. And we also know that one full
turn counterclockwise around from the positive π₯-axis is two π. So this angle will be π plus
two π. And in fact, we could do any
number of full turns around our axis and then an angle of π. Or alternatively, we could
instead take full turns clockwise.

And we know these would instead
be represented by negative angles. And this gives us our first
relationship between polar coordinates. We know the point π one, π
one will be the same as the point π one, π one plus two ππ for any integer
π. This just tells us we can take
π full turns clockwise or counterclockwise before we measure our angle. And it wonβt change the value
of our polar coordinate. In our case, the point π΄ has
polar coordinates π, π. So changing π one to π and π
one to π gives us the following expression.

We know that π, π will be
equal to π, π plus two ππ for any integer π. In other words, we can add or
subtract any even integer multiple of π from our angle. It wonβt change the point. We can see that option (a) does
this. We subtract two π from our
angle of π. In other words, our value of π
is equal to negative one, or weβre taking one extra full turn clockwise around
our axis. We can see that option (b) does
not do this. We would need to choose π
equal to negative one-half. Or alternatively, weβre only
taking half a turn extra clockwise.

We can also see that option (d)
canβt be correct. Our value of π would have to
be three over two. In other words, weβre taking an
extra one and a half turns in the counterclockwise direction. Of course, we canβt use this to
determine whether option (c) or option (e) are correct because both of these
have negative π as their first ordinate.

To determine whether options
(c) and (e) represent point π΄, thereβs one more polar relation we need to
recall. We need to remember that the
point π, π will be the same as the point negative π, π plus two π plus one
times π for any integer π. In fact, we can reason this in
exactly the same way we did before. This time, our first ordinate
is equal to negative π. And when this value is
negative, this just tells us we need to point our line segment ππ΄ in the
opposite direction. But we can then see if we
rotate this vector, π radians, either clockwise or counterclockwise, weβll end
up back where we started. Then just as we did before,
taking full turns clockwise or counterclockwise wonβt change the value of our
polar coordinate.

So this time, when our value of
π was negative, we instead showed you can add or subtract odd integer multiples
of π to our value of π. So we can now use this to
determine whether options (c) and (e) represent our point π΄. In option (c), we can see that
π is negative. However, weβre subtracting an
even integer multiple of point. So option (c) does not
represent the same point. Similarly, in option (e), we
can see that π is negative and weβre adding an even integer multiple of π. So option (e) also canβt be
correct. So in actual fact, out of all
five of these options, only option (a) was the same as our point π΄.

Therefore, given that the point
π΄ had rectangular coordinates negative four, seven, we were able to show the
distance of the point π΄ from the origin which we called π is equal to the
square root of 65. The angle that the line segment
ππ΄ makes with the positive π₯-axis which we called π to two decimal places in
radians was 2.09. And of the five listed options,
only option (a) which was π, π minus two π had the same representation in
polar coordinates as point π΄.