Question

figure abcdef was reflected across the line y = -x to create figure abcdef. what are the coordinates of the pre-image of f? ○ (-2, 4) ○ (4, 2) ○ (2, -4) ○ (-4, -2)

Explanation:

Step1: Find coordinates of \( F' \)

From the graph, \( F' \) has coordinates \( (4, -2) \).

Step2: Apply reflection over \( y = -x \) formula

The rule for reflecting a point \( (x, y) \) over the line \( y = -x \) is \( (x, y) \to (-y, -x) \). Let the pre - image of \( F' \) (i.e., the pre - image of \( F \) is the image of \( F' \) under the reverse reflection, which is the same as applying the reflection formula to \( F' \) to get back to \( F \)). If \( (x', y')=(4, - 2) \) is the image after reflection, and we want to find the pre - image \( (x,y) \), we use the inverse of the reflection rule. The reflection over \( y=-x \) is an involution, meaning applying it twice gives the original point. So if \( (x,y) \) is reflected over \( y = -x \) to get \( (x',y')=(-y,-x) \), then to find \( (x,y) \) from \( (x',y') \), we solve \( x'=-y \) and \( y'=-x \). So \( y=-x' \) and \( x = - y' \).

Substitute \( x' = 4 \) and \( y'=-2 \) into the formulas:
\( x=-y'=-(-2) = 2 \)
\( y=-x'=-4 \)

Wait, no. Wait, the pre - image of \( F \) is the point \( F \) such that when we reflect \( F \) over \( y=-x \), we get \( F' \). Let \( F=(x,y) \), then the reflection of \( F \) over \( y = -x \) is \( F'=(-y,-x) \). We know \( F'=(4,-2) \), so we set up the equations:
\(-y = 4\) and \(-x=-2\)
From \(-y = 4\), we get \( y=-4\)
From \(-x=-2\), we get \( x = 2\)

Wait, no, let's do it correctly. The reflection over \( y=-x \) transforms a point \( (a,b) \) to \( (-b,-a) \). So if \( F=(a,b) \) is the pre - image, and \( F'=(-b,-a) \) is the image. We need to find \( (a,b) \) given \( F'=(4,-2) \). So:
\(-b = 4\) implies \( b=-4\)
\(-a=-2\) implies \( a = 2\)

Wait, but let's check the options. Wait, maybe I made a mistake. Wait, the coordinates of \( F' \): looking at the graph, \( F' \) is at \( (4,-2) \)? Wait, no, let's look at the grid. The x - axis and y - axis: the grid lines are at integers. Let's re - examine the graph. The point \( F' \): from the origin, moving 4 units right on x - axis and 2 units down on y - axis? Wait, no, the y - axis has negative values below the x - axis. Wait, maybe \( F' \) is at \( (4,-2) \)? Wait, no, let's check the options. The options are \( (-2,4) \), \( (4,2) \), \( (2,-4) \), \( (-4,-2) \).

Wait, the correct formula for reflection over \( y=-x \) is: if a point \( (x,y) \) is reflected over \( y=-x \), the image is \( (-y,-x) \). So to find the pre - image, we need to reverse this. Let the image be \( (x',y') \), then the pre - image \( (x,y) \) satisfies \( x'=-y \) and \( y'=-x \), so \( y=-x' \) and \( x=-y' \).

Suppose \( F' \) has coordinates \( (4,-2) \). Then pre - image \( x=-y'=-(-2)=2 \), \( y=-x'=-4 \). So pre - image is \( (2,-4) \)? Wait, no, wait. Wait, maybe I got the image wrong. Wait, the figure \( ABCDEF \) is reflected over \( y = -x \) to get \( A'B'C'D'E'F' \). So \( F' \) is the image of \( F \). So we need to find \( F \) such that reflecting \( F \) over \( y=-x \) gives \( F' \).

Let's find the coordinates of \( F' \) from the graph. Looking at the graph, \( F' \) is at \( (4,-2) \)? Wait, no, the grid: the x - axis is horizontal, y - axis vertical. The point \( F' \): let's count the squares. From the origin (0,0), moving 4 units to the right (x = 4) and 2 units down (y=-2). So \( F'=(4,-2) \).

Now, using the reflection rule over \( y=-x \): \( (x,y)\to(-y,-x) \). So if \( F=(x,y) \) is reflected to \( F'=(-y,-x)=(4,-2) \), then we have the system of equations:
\(-y = 4\) and \(-x=-2\)
Solving \(-y = 4\) gives \( y=-4\)
Solving \(-x=-2\) gives \( x = 2\)
So the pre - image \( F \) has coordinates \( (2,-4) \)? Wait, but let's check the options. Option C is \( (2,-4) \). Wait, but let's verify with another approach. Let's take a point and reflect it over \( y=-x \). For example, if we have a point \( (a,b) \), its reflection over \( y=-x \) is \( (-b,-a) \). Let's take the option \( (2,-4) \). Reflecting \( (2,-4) \) over \( y=-x \): \( -(-4)=4 \), \( -2=-2 \), so the image is \( (4,-2) \), which matches the coordinates of \( F' \) (assuming \( F'=(4,-2) \)). So that works.

Wait, but maybe I misread the coordinates of \( F' \). Let's look again. The graph: \( A' \) is at \( (-2,-2) \)? No, \( A' \) is below the x - axis, \( F' \) is to the right of \( A' \). Wait, maybe \( F' \) is at \( (4,-2) \). So reflecting \( (2,-4) \) over \( y=-x \) gives \( (4,-2) \), which is \( F' \). So the pre - image of \( F \) (since \( F' \) is the image of \( F \)) is \( (2,-4) \).

Answer:

\( (2, - 4) \) (corresponding to the option \( \boldsymbol{(2, -4)} \))