Distances are considered to be non-negative in all cases.
The absolute value function is necessary to get a positive result when the difference of coordinate values is negative.
The square root function is defined to return a positive result always, so no absolute value function is required when the distance is found using the square root function.
No one is going to do all this for 5 points...
Hey there!
1. The graph was flipped vertically and now opens downward.
We know this because there is a negative sign in front of the equation:
2. The graph was moved 3 units to the left.
We know this because there is 3 added to x inside of the parenthesis. Remember, inside (parenthesis) is opposite and acts on x, outside (parenthesis) is same and acts on y.
3. The graph was moved 5 units up.
We know this because there is 5 added at the end of the equation. It acts on y because it is outside of the parenthesis.

Hope this helps!
cot(<em>θ</em>) = cos(<em>θ</em>)/sin(<em>θ</em>)
So if both cot(<em>θ</em>) and cos(<em>θ</em>) are negative, that means sin(<em>θ</em>) must be positive.
Recall that
cot²(<em>θ</em>) + 1 = csc²(<em>θ</em>) = 1/sin²(<em>θ</em>)
so that
sin²(<em>θ</em>) = 1/(cot²(<em>θ</em>) + 1)
sin(<em>θ</em>) = 1 / √(cot²(<em>θ</em>) + 1)
Plug in cot(<em>θ</em>) = -2 and solve for sin(<em>θ</em>) :
sin(<em>θ</em>) = 1 / √((-2)² + 1)
sin(<em>θ</em>) = 1/√(5)