For Exercises 19–24, suppose m is the line with equation x = −5, line n is the line with equation y = 1, line g is the line with equation y = x, and line h is the line with equation y = −2. Given A(9, −3), B(6, 4), and C(−1, −5), what are the coordinates of the vertices of △AʹBʹCʹ for each

The x-axis is: [tex]A' = (9,3)[/tex] [tex]B' =(6,-4)[/tex] [tex]C =(-1,5)[/tex].
The y-axis is [tex]A' = (-9,-3)[/tex] [tex]B' =(-6,4)[/tex] [tex]C =(1,-5)[/tex]

Given

[tex]A = (9,-3)[/tex]

[tex]B = (6,4)[/tex]

[tex]C =(-1,-5)[/tex]

The complete question requires that [tex]\triangle ABC[/tex] be transformed using the following rules of transformation.

[tex](a)\ R_{x\ axis}[/tex]

[tex](b)\ R_{y\ axis}[/tex]

[tex](a)\ R_{x\ axis}[/tex]

The rule of reflection across the x-axis is:

[tex](x,y) \to (x,-y)[/tex]

This means that we negate the y-coordinate of [tex]\triangle ABC[/tex]

So, we have:

[tex]A' = (9,3)[/tex]

[tex]B' =(6,-4)[/tex]

[tex]C =(-1,5)[/tex]

[tex](b)\ R_{y\ axis}[/tex]

The rule of reflection across the y-axis is:

[tex](x,y) \to (-x,y)[/tex]

This means that we negate the x-coordinate of [tex]\triangle ABC[/tex]

So, we have:

[tex]A' = (-9,-3)[/tex]

[tex]B' =(-6,4)[/tex]

[tex]C =(1,-5)[/tex]

Hence, the coordinates of the vertices of [tex]\triangle A'B'C[/tex] after reflection across:

The x-axis is: [tex]A' = (9,3)[/tex] [tex]B' =(6,-4)[/tex] [tex]C =(-1,5)[/tex].
The y-axis is [tex]A' = (-9,-3)[/tex] [tex]B' =(-6,4)[/tex] [tex]C =(1,-5)[/tex]