Solution:
we are given that
Clayton needs to reflect the triangle below across the line y=x.
And as we know the reflection of the point (x,y) across the line y = x is the point (y, x).
Hence the new Traingle will be on the other side of the line y=x and position of x and y-coordinates of the vertices of the trangle gets interchanged.
Hence the Options that applies are:
C’ will move because all points move in a reflection.
The image and the pre-image will be congruent triangles. (Because reflection just changes the position of the trinagle not the property)
The image and pre-image will not have the same orientation because reflections flip figures.
The answer would be
y=(x+4)^2−19
The x is horizontal on a graph and then the y axis is vertical on a graph.
Since only 2 sides are equal (not 3), this is an "isosceles triangle."
