Consider the point on the polar graph below. A. Represent the point's location in the polar form (r,θ) where 0≤r≤5 and −2π≤θ<0.
(r,θ)=

B. Represent the point's location in the polar form (r,θ) where 0≤r≤5 and 0≤θ<2π.
(r,θ)=

C. Represent the point's location in the polar form (r,θ) where −5≤r≤0 and −2π≤θ<0.
(r,θ)=

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a) We can clearly see that the radius is r = 5, now let's find the angle.

We can see that it lies between the angles (2/3)*pi and (5/6)*pi

which one is the angle in between these two?

We can calculate it as:

θ = ( (2/3)*pi + (5/6)*pi)/2 = pi*( 4/6 + 5/6)/2 = pi*(9/6)/2 = pi*(9/12) = pi*(3/4)

But we want to write this in the range:

−2π≤θ<0

Knowing that we have a period of 2*pi

our angle will be equivalent to:

pi*(3/4) - 2*pi = pi*( 3/4 - 2) = pi*(3/4 - 8/4) = pi*(-5/4)

Then this point can be  represented as:

(5, (-5/4)*pi)

B) Same as before, but this time we have 0≤θ<2π

The first value of θ that we found is in that range, so this will be:

(5, (3/4)*pi)

C) Here we have −5≤r≤0 and −2π ≤ θ< 0.

For point a, we already know that:

θ = (-5/4)*pi

But the radius part looks tricky, right?

In the standard notation, the variable r (radius) is defined as:

r = IrI

This means that we always use r as a positive number, and if in the notation we write a negative number, then when we work with that we just change the sign.

This means that having r = -5 is exactly the same as r = 5.

Then we can write this point as:

(-5, (-5/4)*pi)

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

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