Question

which quadrant will δlow be in when it is reflected across the y-axis and then reflected across the x-axis? (1 point) ○ quadrant iv

Explanation:

Step1: Identify original coordinates

First, find the coordinates of \( L \), \( O \), \( W \). From the graph:
- \( L(-5, -2) \)
- \( O(-2, -4) \)
- \( W(-4, -6) \)

Step2: Reflect across y - axis

The rule for reflection across the \( y \) - axis is \( (x,y)\to(-x,y) \).
- For \( L(-5, -2) \): New \( L_1=(5, -2) \)
- For \( O(-2, -4) \): New \( O_1=(2, -4) \)
- For \( W(-4, -6) \): New \( W_1=(4, -6) \)

Step3: Reflect across x - axis

The rule for reflection across the \( x \) - axis is \( (x,y)\to(x,-y) \).
- For \( L_1(5, -2) \): New \( L_2=(5, 2) \)
- For \( O_1(2, -4) \): New \( O_2=(2, 4) \)
- For \( W_1(4, -6) \): New \( W_2=(4, 6) \)

Step4: Determine the quadrant

In the coordinate plane, Quadrant I has \( x>0,y>0 \), Quadrant II has \( x < 0,y>0 \), Quadrant III has \( x < 0,y < 0 \), Quadrant IV has \( x>0,y < 0 \). The new points \( L_2(5,2) \), \( O_2(2,4) \), \( W_2(4,6) \) have \( x>0 \) and \( y>0 \), so they lie in Quadrant I. Wait, but let's re - check the reflection steps. Wait, original triangle is in Quadrant III (since \( x<0,y < 0 \)). Reflect across \( y \) - axis: \( x \) becomes positive, \( y \) remains negative (Quadrant IV). Then reflect across \( x \) - axis: \( y \) becomes positive, \( x \) remains positive. So the final points are in Quadrant I? Wait, maybe I made a mistake in the initial coordinates. Wait, looking at the graph, \( L \) is at \( x=-5,y = - 2 \) (Quadrant III), \( O \) at \( x=-2,y=-4 \) (Quadrant III), \( W \) at \( x = - 4,y=-6 \) (Quadrant III). Reflect across \( y \) - axis: \( (x,y)\to(-x,y) \), so \( L(-5,-2)\to(5,-2) \) (Quadrant IV), \( O(-2,-4)\to(2,-4) \) (Quadrant IV), \( W(-4,-6)\to(4,-6) \) (Quadrant IV). Then reflect across \( x \) - axis: \( (x,y)\to(x,-y) \), so \( (5,-2)\to(5,2) \) (Quadrant I), \( (2,-4)\to(2,4) \) (Quadrant I), \( (4,-6)\to(4,6) \) (Quadrant I). So the triangle after two reflections is in Quadrant I. But the option given is Quadrant IV? Wait, maybe I misread the question. Wait, the question says "reflected across the \( y \) - axis and then reflected across the \( x \) - axis". Wait, maybe the original triangle is in Quadrant III. Let's re - express the reflection rules:

Reflection over \( y \) - axis: \( (x,y)\to(-x,y) \)

Reflection over \( x \) - axis: \( (x,y)\to(x,-y) \)

Composing these two transformations: \( (x,y)\to(-x,y)\to(-x,-y) \). Wait, wait! Oh no, I made a mistake in the second reflection. If we first reflect over \( y \) - axis: \( (x,y)\to(-x,y) \), then reflect over \( x \) - axis: \( (-x,y)\to(-x,-y) \). Oh! I messed up the second reflection. Let's correct that.

Original coordinates:

\( L(-5,-2) \), \( O(-2,-4) \), \( W(-4,-6) \)

First reflection (over \( y \) - axis):

\( L(-5,-2)\to(5,-2) \)

\( O(-2,-4)\to(2,-4) \)

\( W(-4,-6)\to(4,-6) \)

Second reflection (over \( x \) - axis):

For a point \( (a,b) \), reflection over \( x \) - axis is \( (a, - b) \). So:

\( (5,-2)\to(5,2) \)? No, wait, no: reflection over \( x \) - axis changes the sign of \( y \). So \( (5,-2) \) reflected over \( x \) - axis is \( (5, 2) \)? Wait, no, \( y \) - coordinate was \( - 2 \), after reflection over \( x \) - axis, it becomes \( 2 \). But wait, the composition of reflection over \( y \) - axis and then \( x \) - axis is equivalent to a rotation of \( 180^\circ \) about the origin, which is \( (x,y)\to(-x,-y) \). Wait, no: reflection over \( y \) - axis: \( (x,y)\to(-x,y) \); reflection over \( x \) - axis: \( (-x,y)\to(-x,-y) \). So the correct composition is \( (x,y)\to(-x,-y) \). Let's apply that to original points:

For \( L(-5,-2) \): \( -(-5)=5 \), \( -(-2)=2 \), so \( (5,2) \)

For \( O(-2,-4) \): \( -(-2)=2 \), \( -(-4)=4 \), so \( (2,4) \)

For \( W(-4,-6) \): \( -(-4)=4 \), \( -(-6)=6 \), so \( (4,6) \)

These points are in Quadrant I (since \( x>0,y>0 \)). But the option given is Quadrant IV? Wait, maybe the original triangle is in Quadrant III, first reflection over \( y \) - axis takes it to Quadrant IV ( \( x>0,y < 0 \) ), then reflection over \( x \) - axis takes it to Quadrant I ( \( x>0,y>0 \) ). So the answer should be Quadrant I? But the option in the image shows "Quadrant IV" as an option. Wait, maybe I misread the reflection order. If it's reflected over \( x \) - axis first and then \( y \) - axis, but the question says "reflected across the \( y \) - axis and then reflected across the \( x \) - axis".

Wait, let's take a single point. Let's take point \( L(-5,-2) \):

1. Reflect over \( y \) - axis: \( (5,-2) \) (Quadrant IV)
2. Reflect over \( x \) - axis: \( (5,2) \) (Quadrant I)

Point \( O(-2,-4) \):

1. Reflect over \( y \) - axis: \( (2,-4) \) (Quadrant IV)
2. Reflect over \( x \) - axis: \( (2,4) \) (Quadrant I)

Point \( W(-4,-6) \):

1. Reflect over \( y \) - axis: \( (4,-6) \) (Quadrant IV)
2. Reflect over \( x \) - axis: \( (4,6) \) (Quadrant I)

So all three points after two reflections are in Quadrant I. But maybe the question has a typo, or I misread the graph. Wait, maybe the original triangle is in Quadrant II? Wait, no, \( x \) is negative, \( y \) is negative, so Quadrant III.

Wait, maybe the answer is Quadrant I, but the option shown is Quadrant IV. Wait, perhaps I made a mistake in the reflection rule. Let's recall:

- Reflection over \( y \) - axis: \( (x,y)\to(-x,y) \)
- Reflection over \( x \) - axis: \( (x,y)\to(x,-y) \)

So applying to \( L(-5,-2) \):

After \( y \) - axis: \( (5,-2) \)

After \( x \) - axis: \( (5,2) \) (Quadrant I)

Yes, so the triangle is in Quadrant I. But since the option given in the image is "Quadrant IV", maybe there's a mistake, but according to the calculations, after reflecting over \( y \) - axis (Quadrant IV) and then \( x \) - axis (Quadrant I), the answer should be Quadrant I. But perhaps the question was supposed to be reflected over \( x \) - axis first and then \( y \) - axis. Let's check that:

Reflect over \( x \) - axis first: \( (x,y)\to(x,-y) \)

\( L(-5,-2)\to(-5,2) \) (Quadrant II)

Then reflect over \( y \) - axis: \( (-5,2)\to(5,2) \) (Quadrant I). No, same result.

Wait, maybe the original coordinates are different. Let's re - examine the graph:

- \( L \) is at \( x=-5 \), \( y=-2 \) (so \( (-5,-2) \))
- \( O \) is at \( x=-2 \), \( y=-4 \) (so \( (-2,-4) \))
- \( W \) is at \( x=-4 \), \( y=-6 \) (so \( (-4,-6) \))

Yes, that's correct. So after \( y \) - axis reflection: \( (5,-2) \), \( (2,-4) \), \( (4,-6) \) (Quadrant IV)

After \( x \) - axis reflection: \( (5,2) \), \( (2,4) \), \( (4,6) \) (Quadrant I)

So the answer should be Quadrant I. But if the option given is Quadrant IV, maybe there's a mistake in the question or the option. However, based on the calculation, the triangle will be in Quadrant I. But since the user's image shows "Quadrant IV" as an option, maybe I made a mistake. Wait, no, let's check the reflection rules again.

Wait, maybe the question is "reflected across the \( x \) - axis and then across the \( y \) - axis". Let's try that:

Reflect over \( x \) - axis: \( (x,y)\to(x,-y) \)

\( L(-5,-2)\to(-5,2) \) (Quadrant II)

Reflect over \( y \) - axis: \( (-5,2)\to(5,2) \) (Quadrant I). Same result.

Alternatively, maybe the original triangle is in Quadrant II. Wait, if \( L \) was \( (-5,2) \), but in the graph, \( y \) is negative. So no, the \( y \) - coordinates are negative. So the initial triangle is in Quadrant III.

So after two reflections ( \( y \) - axis then \( x \) - axis), it's in Quadrant I.

Answer:

Quadrant I (But if the option given is Quadrant IV, there might be an error in the problem setup or options. However, based on the coordinate transformations, the correct quadrant is Quadrant I.)