Answer:
A(2,2)
Step-by-step explanation:
Let the vertex A has coordinates 
Vectors AB and AB' are perpendicular, then
Vectors AC and AC' are perpendicular, then
Now, solve the system of two equations:
Subtract these two equations:
Substitute it into the first equation:
Then
Rotation by 90° counterclockwise about A(2,2) gives image points B' and C' (see attached diagram)
(x-8) ^ 2 = 121
(x-8) = + / - root (121)
x = 8 +/- root (121)
The solutions are:
x1 = 8 + root (121)
x2 = 8 - root (121)
2a ^ 2 = 8a-6
2a ^ 2-8a + 6 = 0
a ^ 2-4a + 3 = 0
(a-1) (a-3) = 0
The solutions are:
a1 = 1
a2 = 3
x ^ 2 + 12x + 36 = 4
x ^ 2 + 12x + 36-4 = 0
x ^ 2 + 12x + 32 = 0
(x + 4) (x + 8) = 0
The solutions are:
x1 = -8
x2 = -4
x ^ 2-x + 30 = 0
x = (- b +/- root (b ^ 2 - 4 * a * c)) / 2 * a
x = (- (- 1) +/- root ((- 1) ^ 2 - 4 * (1) * (30))) / 2 * (1)
x = (1 +/- root (1 - 120))) / 2
x = (1 +/- root (-119))) / 2
x = (1 +/- root (119) * i)) / 2
The solutions are:
x1 = (1 + root (119) * i)) / 2
x2 = (1 - root (119) * i)) / 2
If you're using a few larger intervals, then your histogram looks more stocky. If you imagine drawing one, it's because you're adding more values into the same category which can make the difference between two intervals much more noticeable. If you're using smaller intervals, however, you can much more accurately assess the difference between two different intervals. For that reason, the transition between one and another interval would look much more 'fluid'.
