Question

triangle abc is rotated clockwise about the origin through 90°. the resulting triangle is then rotated clockwise about the origin through 270°. identify the vertices of the triangle after these transformations.

Explanation:

Step1: Find Original Vertices

From the graph, identify coordinates:

• \( A(-2, 0) \)
• \( B(2, 1) \) (Wait, correction: Looking at grid, \( B \) is at \( (2, 1) \)? Wait, no, recheck: The x-axis grid lines: each square is 1 unit. So \( A \) is at \( (-2, 0) \), \( C \) at \( (-1, 2) \), \( B \) at \( (2, 1) \)? Wait, maybe better to re-express: Let's confirm coordinates:
• \( A \): x=-2, y=0 → \( (-2, 0) \)
• \( B \): x=2, y=1 → \( (2, 1) \)
• \( C \): x=-1, y=2 → \( (-1, 2) \)

Step2: First Rotation (90° Clockwise)

Rule for 90° clockwise rotation about origin: \( (x, y) \to (y, -x) \)

• \( A(-2, 0) \to (0, 2) \)
• \( B(2, 1) \to (1, -2) \)
• \( C(-1, 2) \to (2, 1) \)

Step3: Second Rotation (270° Clockwise)

Rule for 270° clockwise rotation (or 90° counterclockwise): \( (x, y) \to (-y, x) \)
Apply to the result of first rotation:

• \( A_1(0, 2) \to (-2, 0) \) (since \( -y = -2 \), \( x = 0 \)? Wait, no: Wait 270° clockwise rotation rule is \( (x, y) \to (y, -x) \) for 90° clockwise, but 270° clockwise is equivalent to 90° counterclockwise, whose rule is \( (x, y) \to (-y, x) \). Wait, let's recheck rotation rules:
• 90° clockwise: \( (x, y) \to (y, -x) \)
• 180° clockwise: \( (x, y) \to (-x, -y) \)
• 270° clockwise: \( (x, y) \to (-y, x) \) (same as 90° counterclockwise)

Wait, but first rotation is 90° clockwise, then second is 270° clockwise. Let's compute total rotation: 90° + 270° = 360° clockwise. Oh! Wait, 90 + 270 = 360, which is a full rotation. So the net effect is a 360° rotation, which maps each point back to itself. Wait, that's a key insight! 90° + 270° = 360°, so the total rotation is 360°, so the final vertices are the same as original.

Wait, let's verify:
Total rotation angle: 90° + 270° = 360° clockwise. A 360° rotation about origin leaves the figure unchanged. So regardless of the intermediate steps, the final vertices are the original ones.

So original vertices:

• \( A(-2, 0) \)
• \( B(2, 1) \)
• \( C(-1, 2) \)

Wait, but let's confirm with step-by-step:

First rotation (90° clockwise):

• \( A(-2, 0) \to (0, 2) \) (since \( y=0 \), \( -x = 2 \))
• \( B(2, 1) \to (1, -2) \) ( \( y=1 \), \( -x = -2 \))
• \( C(-1, 2) \to (2, 1) \) ( \( y=2 \), \( -x = 1 \))

Second rotation (270° clockwise) on these:

• \( A_1(0, 2) \): apply 270° clockwise rule \( (x, y) \to (-y, x) \) → \( (-2, 0) \)
• \( B_1(1, -2) \): \( (-(-2), 1) = (2, 1) \) (since \( -y = -(-2) = 2 \), \( x = 1 \)? Wait no, 270° clockwise rule: \( (x, y) \to (-y, x) \). So for \( (1, -2) \): \( -y = -(-2) = 2 \), \( x = 1 \)? No, wait \( x=1 \), \( y=-2 \). So \( -y = 2 \), \( x = 1 \)? No, the rule is \( (x, y) \to (-y, x) \), so \( x=1 \), \( y=-2 \) → \( -y = 2 \), \( x = 1 \)? Wait, no, \( (-y, x) \) where \( y=-2 \), so \( -y = 2 \), \( x=1 \) → \( (2, 1) \).
• \( C_1(2, 1) \): \( x=2 \), \( y=1 \) → \( -y = -1 \), \( x=2 \) → \( (-1, 2) \)

Now, let's list the second rotation results:

• \( A_2(-2, 0) \)
• \( B_2(2, 1) \)
• \( C_2(-1, 2) \)

Which are the original coordinates! Because 90° + 270° = 360°, so the total rotation is a full circle, so the triangle returns to its original position.

Answer:

The vertices after the transformations are \( A(-2, 0) \), \( B(2, 1) \), and \( C(-1, 2) \) (same as original, since total rotation is 360°).