Question

a triangle is in the first quadrant of an x-y coordinate system, as shown here:
the triangle is reflected across the x-axis, then it is reflected across the y-axis. which of these transformations will put it back in its original position (with the same orientation it had originally)?
i. a reflection across the line ( y = -x )
ii. a reflection across the y-axis, then a clockwise rotation of ( 90^circ ) about the origin
iii. a reflection across the x-axis, then a reflection across the y-axis
options:

• ii and iii only
• i and iii only
• iii only
• i only

question 14
1 pts
a trapezoid is in the third quadrant of an x-y coordinate system, as shown here:

Explanation:

Step1: Analyze Original Transformations

The original triangle is in the first quadrant. First, it's reflected over the \(x\)-axis (moves to fourth quadrant), then over the \(y\)-axis (moves to third quadrant). Let's denote a point \((x,y)\) in the first quadrant. After \(x\)-axis reflection: \((x, -y)\) (fourth quadrant). After \(y\)-axis reflection: \((-x, -y)\) (third quadrant).

Step2: Analyze Transformation I (Reflect over \(y = -x\))

The reflection of a point \((a,b)\) over \(y=-x\) is \((-b, -a)\). Let's take the final point after original two reflections: \((-x, -y)\). Reflecting over \(y=-x\): \((-(-y), -(-x))=(y, x)\). Wait, no—wait, original point before any transformations is \((x,y)\) (first quadrant). Wait, maybe better to track the original point. Wait, the question is: which transformation will put the triangle (after the two reflections: \(x\)-axis then \(y\)-axis, resulting in third quadrant) back to original (first quadrant) with same orientation. Wait, no—the question is: the triangle is first reflected over \(x\)-axis, then \(y\)-axis (so it's in third quadrant). Now, we need a transformation that takes it from third quadrant back to first quadrant with same orientation.

Wait, let's re-express:

Original position: first quadrant, point \(P=(x,y)\), \(x>0\), \(y>0\).

After reflection over \(x\)-axis: \(P_1=(x, -y)\) (fourth quadrant, \(y<0\), \(x>0\)).

After reflection over \(y\)-axis: \(P_2=(-x, -y)\) (third quadrant, \(x<0\), \(y<0\)).

Now, we need to find which of I, II, III will take \(P_2\) back to \(P=(x,y)\) with same orientation.

Step3: Analyze Transformation I (Reflect \(P_2\) over \(y=-x\))

Reflection of \(P_2=(-x, -y)\) over \(y=-x\): the formula is \((-b, -a)\) for point \((a,b)\). So \(a=-x\), \(b=-y\). So reflection is \((-(-y), -(-x))=(y, x)\). Wait, that's not \((x,y)\). Wait, maybe I messed up. Wait, reflection over \(y=-x\) is \((x,y) \to (-y, -x)\). So if we have a point in third quadrant: \((-x, -y)\) (where \(x>0\), \(y>0\)), reflecting over \(y=-x\): \((-(-y), -(-x))=(y, x)\). Wait, no—wait, let's take a concrete example. Let \(x=1\), \(y=1\) (point (1,1) in first quadrant). After \(x\)-axis reflection: (1, -1) (fourth quadrant). After \(y\)-axis reflection: (-1, -1) (third quadrant). Now, reflect (-1, -1) over \(y=-x\): the reflection of (a,b) over \(y=-x\) is (-b, -a). So (a,b)=(-1, -1), so (-(-1), -(-1))=(1,1). Oh! Wait, that works. So (1,1) reflected over \(y=-x\) is (1,1)? Wait, no—wait, (1,1) reflected over \(y=-x\) is (-1, -1). Wait, I had the formula reversed. The correct reflection over \(y=-x\) is: for a point \((x,y)\), the image is \((-y, -x)\). So (1,1) reflected over \(y=-x\) is (-1, -1). So to reverse that, reflecting (-1, -1) over \(y=-x\) gives (1,1). Yes! Because applying reflection over \(y=-x\) twice gives the original point. So if we have \(P_2=(-1, -1)\) (third quadrant), reflecting over \(y=-x\) gives (1,1) (first quadrant), which is original. So Transformation I works.

Step4: Analyze Transformation II (Reflect over \(y\)-axis, then clockwise 90° rotation)

First, reflect \(P_2=(-x, -y)\) over \(y\)-axis: the reflection of \((-x, -y)\) over \(y\)-axis is \((x, -y)\) (fourth quadrant). Then, a clockwise 90° rotation about the origin. The rotation of a point \((a,b)\) clockwise 90° is \((b, -a)\). So \((x, -y)\) rotated clockwise 90°: \((-y, -x)\). Which is not \((x,y)\). Wait, let's take the concrete point (-1, -1). Reflect over \(y\)-axis: (1, -1) (fourth quadrant). Rotate clockwise 90°: the formula for clockwise 90° rotation is \((x,y) \to (y, -x)\). So (1, -1) becomes (-1, -1). Wait, that's back to \(P_2\), not original. Wait, maybe I made a mistake. Wait, clockwise 90° rotation: the standard matrix is \(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) for counterclockwise, so clockwise is \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\)? Wait, no—let's recall: rotating a point \((x,y)\) clockwise 90° around the origin gives \((y, -x)\). So (1, -1) rotated clockwise 90°: \(y=-1\), \(-x=-1\), so (-1, -1). So that's not original. So Transformation II does not work.

Step5: Analyze Transformation III (Reflect over \(x\)-axis, then \(y\)-axis)

Wait, the original two reflections were \(x\)-axis then \(y\)-axis, resulting in \(P_2=(-x, -y)\). Now, Transformation III is: reflect \(P_2\) over \(x\)-axis, then over \(y\)-axis. Let's do that. Reflect \(P_2=(-x, -y)\) over \(x\)-axis: \((-x, y)\) (second quadrant). Then reflect over \(y\)-axis: \((x, y)\) (first quadrant). Perfect! So that's the original point. So Transformation III works.

Wait, but earlier when we analyzed Transformation II, we saw it didn't work. Wait, but let's recheck Transformation II. Wait, the original problem's Transformation II is: "a reflection across the \(y\)-axis, then a clockwise rotation of \(90^\circ\) about the origin". Wait, maybe I messed up the order. Wait, the triangle is currently at \(P_2=(-x, -y)\) (after original two reflections). Transformation II is: first reflect over \(y\)-axis, then rotate clockwise 90°. So \(P_2=(-x, -y)\) reflected over \(y\)-axis: \((x, -y)\) (fourth quadrant). Then rotate clockwise 90°: \((-y, -x)\). Wait, but if we take the concrete point (-1, -1): reflect over \(y\)-axis: (1, -1). Rotate clockwise 90°: ( -1, -1)? Wait, no—clockwise 90° rotation of (1, -1): the formula is (x,y) → (y, -x). So (1, -1) → (-1, -1). Which is \(P_2\), not original. So Transformation II does not work.

Wait, but earlier when we did Transformation I: reflecting \(P_2=(-1, -1)\) over \(y=-x\) gives (1,1), which is original. Transformation III: reflecting \(P_2\) over \(x\)-axis (to (-1, 1)) then over \(y\)-axis (to (1,1)), which is original. So I and III work.

Wait, but let's check the options. The options are:

- II and III only

- I and III only

- III only

- I only

So according to our analysis, I (reflection over \(y=-x\)) works, III (reflect over \(x\)-axis then \(y\)-axis) works, II does not. So the correct option is "I and III only".

Answer:

I and III only (the option with "I and III only")