Question

9
how many degrees has △abc been rotated counterclockwise about the origin?

a. 270°
b. 180°
c. 360°
d. 90°

Explanation:

Step1: Identify coordinates of a vertex

Take point \( A \) from \( \triangle ABC \). Let's find its coordinates. From the graph, \( A \) is at \( (1, 5) \) (wait, no, looking at the grid: let's check the grid lines. Wait, the original \( A \): x - coordinate, each grid is 1 unit. Let's see, original \( A \): x is 1? Wait, no, the y - axis is at x = 0. Let's re - check. Original \( A \): looking at the black triangle, \( A \) is at (1, 5)? Wait, no, the grid: from y - axis (x = 0), moving right 1, 2, 3? Wait, maybe better to take a point. Let's take point \( A \): original \( A \) is at (1, 5)? Wait, no, the red triangle \( A' \) is at (5, - 2)? Wait, no, let's do it properly. Let's take point \( A \) in \( \triangle ABC \): let's say \( A=(1,5) \)? Wait, no, looking at the graph, original \( A \): x - coordinate is 1 (since from y - axis, 1 unit right), y - coordinate is 5 (5 units up). Then \( A' \) is at (5, - 1)? Wait, no, maybe I made a mistake. Alternatively, use the rotation rules.

Rotation rules:
- 90° counterclockwise: \( (x,y)\to(-y,x) \)
- 180° counterclockwise: \( (x,y)\to(-x,-y) \)
- 270° counterclockwise: \( (x,y)\to(y,-x) \)
- 360° counterclockwise: \( (x,y)\to(x,y) \)

Let's take a point from \( \triangle ABC \), say \( A \). Let's find its coordinates. Looking at the black triangle, \( A \) is at (1, 5)? Wait, no, the grid: each square is 1 unit. Let's look at the x - axis and y - axis. The original \( A \): x = 1 (since from y - axis (x = 0), 1 unit to the right), y = 5 (5 units up). Then \( A' \) in the red triangle: let's see, \( A' \) is at (5, - 1)? Wait, no, maybe I messed up. Wait, another approach: let's take point \( B \) in \( \triangle ABC \): \( B=(4,8) \). Then \( B' \) in \( \triangle A'B'C' \): \( B'=(8, - 4) \)? Wait, no, let's check the rotation rule for 90° counterclockwise: \( (x,y)\to(-y,x) \). For \( (4,8) \), 90° counterclockwise would be \( (-8,4) \), which is not \( B' \). For 180° counterclockwise: \( (x,y)\to(-x,-y) \). For \( (4,8) \), 180° would be \( (-4,-8) \), not \( B' \). For 270° counterclockwise: \( (x,y)\to(y,-x) \). For \( (4,8) \), 270° counterclockwise: \( (8, - 4) \). Wait, but let's check \( A \). Original \( A \): let's say \( A=(1,5) \). 270° counterclockwise: \( (5, - 1) \). Wait, maybe my coordinate for \( A \) is wrong. Let's take \( A \) as (1, 5), \( A' \) as (5, - 1). Wait, no, maybe the original \( A \) is (1, 5) and \( A' \) is (5, - 1). Wait, 270° rotation: \( (x,y)\to(y,-x) \). So if \( x = 1 \), \( y = 5 \), then \( (5, - 1) \), which matches? Wait, no, maybe I got the coordinates wrong. Alternatively, let's take a point with integer coordinates. Let's take point \( C \) in \( \triangle ABC \): \( C=(7,5) \). Then 270° counterclockwise rotation: \( (5, - 7) \). But \( C' \) in the red triangle: let's see, \( C' \) is at (5, - 8)? Wait, maybe I'm overcomplicating. Let's use the visual. A 90° counterclockwise rotation would turn the triangle 90° around the origin. A 180° rotation would flip it 180°, making the triangle opposite. A 270° rotation is 3/4 of a circle. Wait, the red triangle is a 90° counterclockwise rotation? Wait, no, the options are 270, 180, 360, 90. Wait, let's check the rotation of a point. Let's take point \( A \): original \( A \) (let's find correct coordinates). Looking at the grid, original \( A \): x - coordinate is 1 (since from y - axis, 1 unit right), y - coordinate is 5 (5 units up). \( A' \) is at (5, - 1). Wait, 90° counterclockwise: \( (x,y)\to(-y,x) \). For \( (1,5) \), \( (-5,1) \), which is not \( A' \). 180°: \( (-1,-5) \), not. 270°: \( (5, - 1) \), which matches. Wait, yes! So \( (x,y)=(1,5) \), 270° counterclockwise: \( (y,-x)=(5, - 1) \), which is \( A' \). So the rotation is 270°? Wait, no, wait the option A is 270°, B is 180°, C is 360°, D is 90°. Wait, maybe I made a mistake. Wait, let's take another point. Original \( B \): \( (4,8) \). 270° counterclockwise: \( (8, - 4) \). Let's check \( B' \): is \( B' \) at (8, - 4)? Looking at the red triangle, \( B' \) is at (8, - 4)? Yes! So \( B=(4,8) \), after 270° counterclockwise rotation, \( (8, - 4) \), which is \( B' \). So the rotation is 270°? Wait, no, the option A is 270°, but wait, maybe I messed up the rotation direction. Wait, counterclockwise 270° is the same as clockwise 90°. Wait, let's check the rotation rule again. The formula for 90° counterclockwise rotation about the origin is \( (x,y)\to(-y,x) \), 180° is \( (x,y)\to(-x,-y) \), 270° counterclockwise is \( (x,y)\to(y,-x) \), 360° is \( (x,y)\to(x,y) \). So for point \( B=(4,8) \), applying 270° counterclockwise: \( (8, - 4) \), which is \( B' \). So the triangle has been rotated 270° counterclockwise? Wait, but let's check the answer options. Option A is 270°, B is 180°, C is 360°, D is 90°. Wait, maybe I made a mistake in coordinates. Let's re - check the coordinates. Original \( A \): looking at the black triangle, \( A \) is at (1, 5)? Wait, no, the x - axis: from y - axis (x = 0), moving right 1, 2, 3, 4, 5? Wait, maybe the original \( A \) is at (1, 5), \( A' \) is at (5, - 1). So \( (x = 1,y = 5)\to(y = 5,-x=-1) \), which is \( (5, - 1) \), which is 270° counterclockwise. So the rotation is 270°, so the answer is A.

Wait, no, wait the problem is asking how many degrees has \( \triangle ABC \) been rotated counterclockwise about the origin. Let's take a point, say \( A \) in \( \triangle ABC \): let's assume \( A=(1,5) \), \( A'=(5,-1) \). Using the rotation formula for 270° counterclockwise: \( (x,y)\to(y, - x) \). So \( x = 1 \), \( y = 5 \), so \( (5, - 1) \), which matches \( A' \). So the rotation is 270°, so the correct option is A.

Step1: Select a point from \( \triangle ABC \)

Let's choose point \( B \) in \( \triangle ABC \). From the graph, the coordinates of \( B \) are \( (4,8) \).

Step2: Apply rotation rules

- For a 90° counterclockwise rotation about the origin, the transformation rule is \( (x,y)\to(-y,x) \). Applying this to \( (4,8) \), we get \( (-8,4) \), which does not match the coordinates of \( B' \) (the image of \( B \) in \( \triangle A'B'C' \)).
- For a 180° counterclockwise rotation about the origin, the transformation rule is \( (x,y)\to(-x,-y) \). Applying this to \( (4,8) \), we get \( (-4,-8) \), which does not match the coordinates of \( B' \).
- For a 270° counterclockwise rotation about the origin, the transformation rule is \( (x,y)\to(y, - x) \). Applying this to \( (4,8) \), we get \( (8,-4) \), which matches the coordinates of \( B' \) (by visually inspecting the graph, \( B' \) has coordinates that are consistent with \( (8,-4) \)).
- For a 360° counterclockwise rotation about the origin, the transformation rule is \( (x,y)\to(x,y) \). Applying this to \( (4,8) \), we get \( (4,8) \), which does not match the coordinates of \( B' \).

Answer:

A. \( 270^\circ \)