Which statement explains why △abc is congruent to △a′b′c′ ? you can map △abc onto △a′b′c′ reflecting it across the line y = x and rotating it 90° counterclockwise about the origin, which is a sequence of rigid motions. you can map △abc onto △a′b′c′ by reflecting it across the x-axis and then across the y-axis, which is a sequence of rigid motions. you can map △abc onto △a′b′c′ by translating it 2 units up and reflecting it across the y-axis, which is a sequence of rigid motions. you can map △abc onto △a′b′c′ by translating it 6 units left and reflecting it over the x-axis, which is a sequence of rigid motions. two triangles on the coordinate plane. triangle a b c has vertex a at 1 comma negative 3, vertex b at 5 comma 3, and vertex c at 4 comma negative 1. triangle a prime b prime c prime has vertex a prime at negative 5 comma 3, vertex b prime at negative 1 comma negative 3, and vertex c prime at negative 2 comma 1.

The correct answer is:

You can map ABC to A'B'C' by translating it 6 units left and reflecting it across the x-axis, which is a series of rigid motions.

Explanation:

In ABC, the coordinates are:
A(1, -3)
B(5, 3)
C(4, -1).

In A'B'C', the coordinates are:
A'(-5, 3)
B'(-1, -3)
C'(-2, 1)

Each point is mapped to its image:
A(1, -3)→A'(-5, 3)
B(5, 3)→B'(-1, -3)
C(4, -1)→C'(-2, 1)

Comparing the x-coordinates in the pre-image and image, we notice that the image has an x-coordinate that is 6 less than that of the pre-image:
1-6 = -5
5-6 = -1
4-6 = -2

This means that the figure must be translated 6 units left; that is the only way to have this change on the pre-image to form the image.

Comparing the y-coordinates of the pre-image with those of the image, we notice that they are negated:
-(-3) = 3
-(3) = -3
-(-1) = 1

This means the pre-image was reflected across the x-axis; this is the only way to negate the y-coordinate and not change the x-coordinate.

These are rigid motions because they do not change the shape or size, they simply move it and change its orientation.

Hey mate where is the graph