Vertical angles 2 and 5, 1 and 6, 3 and 8, 7 and 4
linear pair 1 and 5, 5 and 6, 6 and 2, 2 and 1, and also 3 and 7, 7 and 8, 8 and 4, 4 and 3
congruent angles - all vertical pairs are congruent angles
2 congruent to 5, 1 congruent to 6, 3 congruent to 8, 7 congruent to 4

What's The Fixed Point? Well, Let's Assume X=0, this is our point.

OR

I'm unaware of the nature of the question, so here are some different ways a fixed point is found based on the merits of the question.
A(x - x₁)(x - x₂) = 0
a=5, x₁=-3, x₂ = 3
so:
5(x-(-3))(x-3)=0
5(x+3)(x-3)=0
5(x²-9) = 0
5x² - 45 = 0 ← answer
Since there is nothing on the left side of the equation besides the absolute value, you have to take 6m out of the absolute value and create 2 separate equations. These 2 equation will be 6m = 42 and 6m = -42.
You solve the equations normally:
6m = 42 6m = -42 (inverse to get m by itself, do on both sides)
/6 /6 /6 /6
m = 7 m = -7
The answer is a solution set:
[7, -7]