Reflection Over Parallel Lines Please complete the attached reflection. Thanks!

A(3, -5)

B(6, -2)

C(9, -2)

Step-by-step explanation:

If we have a point (x, y), and we do a reflection over the axis y = a, then the only thing that will change in our point is the value of x.

Now, the distance between x and a must remain constant before and after the reflection.

so if x - a = d

then the new position of the point will be:

(a - d, y) = (2a - x, y).

I will use that relationship for the 3 points

A)

We start with the point (1, -5)

The reflection over y = -1 leaves.

The distance between 1 and -1 is = 1 - (-1) = 2.

Then the new point is (-1 - 2, -5) = (-3, -5)

Now we do a reflection over y = 1, so D = -3 - 1 = -2

Then the new point is:

A = (1 -(-2), -5) = (3, -5)

B) (2, -2)

Reflection over y = -1.

distance, d = 2 - (-1) = 3

the point is (-1 - 3, -2) = (-4, -2)

Now, a reflection over y = 1.

The distance is D = -4 - 1 = -5

The new point is (1 - (-5), 2) = (6, -2)

C) (5, -2)

reflection over y = -1

Distance: D = 5 - ( - 1) = 6

New point: (-1 - 6, -2) = (-7, -2)

Reflection over y = 1.

Distance D = -7 - 1 = -8

New point ( 1 - (-8), -2) = (9, -2)

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.