Question

What single transformation maps ∆abc onto ∆a'b'c'? graph shows 2 triangles plotted on a coordinate plane. triangle 1 is at a(minus 3, 1), b(-1, 2), c(- 2, 1). triangle 2 is at a prime (-1, -3), b prime (-2,-1), c prime (-1,-2). a. rotation 90° clockwise about the origin b. rotation 90° counterclockwise about the origin c. reflection across the x-axis d. reflection across the line y = x

Answer 1

The correct transformation mapping ∆abc onto ∆a'b'c' is a rotation 90° clockwise about the origin. The coordinates transformation rules confirm that each vertex of ∆abc is rotated accordingly to match the vertices of ∆a'b'c'.

To determine the transformation that maps ∆ABC onto ∆A'B'C', let's analyze their relative positions and orientations based on their coordinates.

Given:

Triangle 1: A(-3, 1), B(-1, 2), C(-2, 1)

Triangle 2: A'(-1, -3), B'(-2, -1), C'(-1, -2)

Let's evaluate the transformations:

a. Rotation 90° clockwise about the origin would change the signs of x-coordinates and swap x and y values. This doesn't match the corresponding points from ∆ABC to ∆A'B'C'.

b. Rotation 90° counterclockwise about the origin would again change the signs of x-coordinates and swap x and y values, which also doesn't align the points.

c. Reflection across the x-axis would only change the signs of y-coordinates, but the x-coordinates would remain the same, and that doesn't match the corresponding points.

d. Reflection across the line y = x swaps the x and y values. Let's apply this transformation to see if it maps ∆ABC onto ∆A'B'C':

For ∆ABC:

- A(-3, 1) → A'(1, -3)

- B(-1, 2) → B'(2, -1)

- C(-2, 1) → C'(1, -2)

These new coordinates indeed match the coordinates of ∆A'B'C', suggesting that the transformation is a reflection across the line y = x, making the correct answer: d. Reflection across the line y = x.

Graph from question: