Question

what single transformation maps δabc onto δabc? graph of coordinate plane with triangle abc and triangle abc and options: 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

Explanation:

Step1: Identify Coordinates

Find coordinates of \( \triangle ABC \) and \( \triangle A'B'C' \). Let's assume:
- \( A(-3, 1) \), \( B(-1, 2) \), \( C(-2, 1) \)
- \( A'(-1, -3) \), \( B'(-2, -1) \), \( C'(-1, -2) \)

Step2: Test Rotation Rules

Recall rotation rules:
- \( 90^\circ \) clockwise: \( (x, y) \to (y, -x) \)
- \( 90^\circ \) counterclockwise: \( (x, y) \to (-y, x) \)

Test \( A(-3, 1) \) with \( 90^\circ \) clockwise: \( (1, 3) eq A'(-1, -3) \). Wait, maybe miscalculation. Wait, let's re-express coordinates. Wait, maybe I mixed up. Let's take \( A(-3, 1) \). For \( 90^\circ \) clockwise about origin: \( (x,y) \to (y, -x) \). So \( (-3,1) \to (1, 3) \)? No, that's not matching. Wait, maybe \( 90^\circ \) clockwise is \( (x,y) \to (y, -x) \), but let's check \( B(-1,2) \). \( 90^\circ \) clockwise: \( (2, 1) \)? No. Wait, maybe I got the coordinates wrong. Wait, looking at the graph:

Original \( A \) is at (-3,1), \( B \) at (-1,2), \( C \) at (-2,1).

Transformed \( A' \) is at (-1, -3), \( B' \) at (-2, -1), \( C' \) at (-1, -2).

Wait, let's apply \( 90^\circ \) clockwise rotation: \( (x,y) \to (y, -x) \).

For \( A(-3,1) \): \( (1, 3) \)? No. Wait, maybe \( 90^\circ \) counterclockwise: \( (x,y) \to (-y, x) \).

For \( A(-3,1) \): \( (-1, -3) \). Yes! \( -y = -1 \), \( x = -3 \)? Wait, no: \( (-y, x) \) for \( (-3,1) \) is \( (-1, -3) \). Yes! \( -y = -1 \), \( x = -3 \)? Wait, \( y = 1 \), so \( -y = -1 \), \( x = -3 \). So \( (-1, -3) \), which is \( A' \).

Check \( B(-1,2) \): \( -y = -2 \), \( x = -1 \)? Wait, no: \( (-y, x) \) is \( (-2, -1) \), which is \( B' \). Yes! \( B(-1,2) \to (-2, -1) = B' \).

Check \( C(-2,1) \): \( -y = -1 \), \( x = -2 \)? Wait, \( (-y, x) \) is \( (-1, -2) \), which is \( C' \). Yes! So \( 90^\circ \) counterclockwise rotation about origin.

Wait, but option A is \( 90^\circ \) clockwise, B is \( 90^\circ \) counterclockwise. Wait, my calculation for \( 90^\circ \) counterclockwise: \( (x,y) \to (-y, x) \). Let's confirm:

Original \( A(-3,1) \): \( -y = -1 \), \( x = -3 \)? No, \( x = -3 \), so \( (-y, x) = (-1, -3) \), which matches \( A'(-1, -3) \).

Original \( B(-1,2) \): \( -y = -2 \), \( x = -1 \)? Wait, \( x = -1 \), so \( (-2, -1) \), which matches \( B'(-2, -1) \).

Original \( C(-2,1) \): \( -y = -1 \), \( x = -2 \)? Wait, \( x = -2 \), so \( (-1, -2) \), which matches \( C'(-1, -2) \). Yes! So that's \( 90^\circ \) counterclockwise? Wait, no: \( 90^\circ \) counterclockwise rotation formula is \( (x,y) \to (-y, x) \). Let's verify with a point. Take \( (1,0) \), \( 90^\circ \) counterclockwise should be \( (0,1) \). Using formula: \( -y = 0 \), \( x = 1 \)? No, wait, no: \( (x,y) = (1,0) \), \( -y = 0 \), \( x = 1 \)? No, that's \( (0,1) \)? Wait, I think I mixed up the formula. Correct \( 90^\circ \) counterclockwise: \( (x,y) \to (-y, x) \). So \( (1,0) \to (0,1) \), which is correct. Wait, \( (0,1) \) rotated \( 90^\circ \) counterclockwise is \( (-1,0) \), using formula: \( -y = -1 \), \( x = 0 \)? No, \( (0,1) \to (-1, 0) \), which is correct. So formula is correct.

So for \( A(-3,1) \): \( -y = -1 \), \( x = -3 \)? Wait, no: \( x = -3 \), \( y = 1 \), so \( -y = -1 \), \( x = -3 \)? No, the formula is \( (x,y) \to (-y, x) \), so \( x=-3 \), \( y=1 \), so \( -y = -1 \), \( x = -3 \)? Wait, no, \( x \) is -3, so the new x is -y = -1, new y is x = -3. So \( (-1, -3) \), which is \( A' \). Yes! That matches.

For \( B(-1,2) \): \( x=-1 \), \( y=2 \), so \( -y = -2 \), \( x = -1 \)? Wait, no, new x is -y = -2, new y is x = -1. So \( (-2, -1) \), which is \( B' \). Perfect.

For \( C(-2,1) \): \( x=-2 \), \( y=1 \), so \( -y = -1 \), \( x = -2 \)? No, new x is -y = -1, new y is x = -2. So \( (-1, -2) \), which is \( C' \). Perfect. So this is \( 90^\circ \) counterclockwise rotation about the origin? Wait, no, wait: \( 90^\circ \) counterclockwise is \( (x,y) \to (-y, x) \), which we applied, and it works. Wait, but let's check \( 90^\circ \) clockwise: \( (x,y) \to (y, -x) \). For \( A(-3,1) \): \( (1, 3) \), which is not \( A' \). So the correct transformation is \( 90^\circ \) counterclockwise about the origin? Wait, but option B is "rotation \( 90^\circ \) counterclockwise about the origin", option A is clockwise. Wait, but according to the calculation, it's \( 90^\circ \) counterclockwise? Wait, no, wait: when we rotate \( 90^\circ \) counterclockwise, the formula is \( (x,y) \to (-y, x) \), which we used, and it worked. So the correct answer is B? Wait, but let's check reflection.

Reflection across x-axis: \( (x,y) \to (x, -y) \). For \( A(-3,1) \): \( (-3, -1) eq A'(-1, -3) \). So C is out.

Reflection across \( y=x \): \( (x,y) \to (y,x) \). For \( A(-3,1) \): \( (1, -3) eq A'(-1, -3) \). So D is out.

Wait, but earlier when I thought \( 90^\circ \) clockwise, I made a mistake. The correct formula for \( 90^\circ \) clockwise is \( (x,y) \to (y, -x) \). Let's test that. For \( A(-3,1) \): \( (1, 3) eq A' \). So \( 90^\circ \) counterclockwise works. So the transformation is \( 90^\circ \) counterclockwise about the origin, which is option B.

Wait, but wait, maybe I mixed up clockwise and counterclockwise. Let's recall: rotating a point \( 90^\circ \) counterclockwise around the origin, the x and y coordinates switch, and the new x is the negative of the original y. So \( (x,y) \to (-y, x) \). Rotating \( 90^\circ \) clockwise, it's \( (x,y) \to (y, -x) \). So in our case, the transformation is \( 90^\circ \) counterclockwise, so option B.

Answer:

B. rotation \( 90^\circ \) counterclockwise about the origin