Reflection Across Y=-x: Which Point Maps To Itself?

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Hey math whizzes! Ever wondered what happens when you reflect a point across a specific line? Today, we're diving deep into the world of coordinate geometry to tackle a super cool problem: identifying which point maps onto itself after a reflection across the line $y=-x$. This isn't just about memorizing formulas, guys; it's about understanding the why behind the transformations. We'll break down what reflection means in this context, how the line $y=-x$ plays a role, and then systematically test each of the given points to find our special mapping. So, grab your pencils, dust off those graph papers, and let's get this mathematical party started!

Understanding Reflection Across the Line $y=-x$

Alright, so what exactly does it mean to reflect a point across a line? Imagine the line $y=-x$ as a mirror. When you reflect a point, you're essentially finding a new point on the opposite side of the mirror such that the line segment connecting the original point and its reflection is perpendicular to the line of reflection, and the line of reflection bisects this segment. In simpler terms, it's like looking at yourself in a mirror – the image you see is the reflection.

Now, let's talk specifically about the line $y=-x$. This line is pretty special. It passes through the origin (0,0) and has a slope of -1. It essentially cuts through the second and fourth quadrants. When we reflect a point $(x, y)$ across the line $y=-x$, something really neat happens to its coordinates. The rule is quite straightforward: the reflected point $(x', y')$ will have coordinates $(y', x') = (-y, -x)$. See that? The x and y coordinates swap places, and both of them get negated. It's like a double transformation – a swap followed by a sign change. This rule is derived from the geometric properties of perpendicularity and bisection we talked about. Let's quickly prove this: Suppose we have a point $P(x,y)$ and its reflection $P'(x',y')$. The midpoint of $PP'$ is $\left(\frac{x+x'}{2}, \frac{y+y'}{2}\right)$. This midpoint must lie on the line $y=-x$, so $\frac{y+y'}{2} = -\frac{x+x'}{2}$, which simplifies to $y+y' = -x-x'$. The slope of the line segment $PP'$ is $\frac{y'-y}{x'-x}$. Since $PP'$ must be perpendicular to $y=-x$ (which has a slope of -1), the slope of $PP'$ must be 1 (because the product of slopes of perpendicular lines is -1). So, $\frac{y'-y}{x'-x} = 1$, which means $y'-y = x'-x$, or $y'-x' = y-x$. Now we have two equations:

  1. y+y′=−x−x′y+y' = -x-x'

  2. y′−x′=y−xy'-x' = y-x

From equation 2, we get $y' = y-x+x'$. Substitute this into equation 1:

y+(y−x+x′)=−x−x′y + (y-x+x') = -x-x'

2y−x+x′=−x−x′2y - x + x' = -x - x'

2y+x′=−x′2y + x' = -x'

2y=−2x′2y = -2x'

x′=−yx' = -y

Now substitute $x' = -y$ back into $y'-x' = y-x$:

y′−(−y)=y−xy' - (-y) = y-x

y′+y=y−xy' + y = y-x

y′=−xy' = -x

So, the reflected point is indeed $(x', y') = (-y, -x)$. Pretty cool, right? Understanding this transformation rule is key to solving our problem.

Finding the Point That Maps to Itself

So, we're looking for a point $(x, y)$ that, after being reflected across the line $y=-x$, ends up being the exact same point. Using our transformation rule $(x', y') = (-y, -x)$, this means we need to find a point $(x, y)$ where $(x, y) = (-y, -x)$.

This equality gives us two conditions that must be met simultaneously:

  1. x=−yx = -y

  2. y=−xy = -x

Notice something? These two conditions are actually the same equation, just rearranged! If $x = -y$, then multiplying both sides by -1 gives $-x = y$, which is our second condition. Similarly, if $y = -x$, then multiplying by -1 gives $-y = x$, which is our first condition. So, we just need to satisfy one of these conditions: $x = -y$, or equivalently, $y = -x$.

What does this condition $y = -x$ mean geometrically? It means the point $(x, y)$ must lie on the line of reflection itself! If a point is already on the mirror line, reflecting it doesn't move it at all – it stays exactly where it is. Think about it: if you're standing in front of a mirror, and you're standing on the mirror, your reflection is just... you. So, any point that lies on the line $y=-x$ will map onto itself after a reflection across $y=-x$. This is a fundamental concept in reflections: points on the line of reflection are invariant under that reflection.

Testing the Given Points

Now for the fun part – putting our knowledge to the test with the points provided! We need to check which of these points satisfies the condition $y = -x$. Let's go through them one by one.

Point 1: $(-4, -4)$

For this point, $x = -4$ and $y = -4$. Let's check if $y = -x$.

  • Is $-4 = -(-4)$?

  • -4 = 4$?

This is false. So, the point $(-4, -4)$ does not lie on the line $y=-x$. Let's see where it maps after reflection. Using our rule $(x', y') = (-y, -x)$, the reflection of $(-4, -4)$ is $(-(-4), -(-4)) = (4, 4)$. Clearly, $(-4, -4)$ maps to $(4, 4)$, not itself.

Point 2: $(-4, 0)$

Here, $x = -4$ and $y = 0$. Let's check if $y = -x$.

  • Is $0 = -(-4)$?

  • 0 = 4$?

This is also false. The point $(-4, 0)$ does not lie on $y=-x$. Let's find its reflection: $(x', y') = (-y, -x) = (-0, -(-4)) = (0, 4)$. So, $(-4, 0)$ maps to $(0, 4)$. Nope, not mapping to itself.

Point 3: $(0, -4)$

In this case, $x = 0$ and $y = -4$. Let's check our condition $y = -x$.

  • Is $-4 = -(0)$?

  • -4 = 0$?

This is false. The point $(0, -4)$ is not on the line $y=-x$. Its reflection is $(x', y') = (-y, -x) = (-(-4), -(0)) = (4, 0)$. So, $(0, -4)$ maps to $(4, 0)$. Still not the one we're looking for.

Point 4: $(4, -4)$

Finally, let's examine the point $(4, -4)$. Here, $x = 4$ and $y = -4$. Let's apply the condition $y = -x$.

  • Is $-4 = -(4)$?

  • -4 = -4$?

True! This statement is correct. The point $(4, -4)$ satisfies the condition $y = -x$. This means the point $(4, -4)$ lies on the line $y=-x$. Therefore, when reflected across $y=-x$, this point will map onto itself. Let's verify this using the reflection rule $(x', y') = (-y, -x)$. For $(4, -4)$, the reflection is $(x', y') = (-(-4), -(4)) = (4, -4)$. And voilà! The reflected point is the same as the original point.

Conclusion

So, there you have it, math enthusiasts! After carefully examining the properties of reflection across the line $y=-x$ and applying the transformation rule $(x, y) \rightarrow (-y, -x)$, we found that a point maps onto itself if and only if it lies on the line of reflection. By testing each of the given points $( -4, -4)$, $( -4, 0)$, $(0, -4)$, and $(4, -4)$, we discovered that only $(4, -4)$ satisfies the condition $y = -x$. Therefore, the point that would map onto itself after a reflection across the line $y=-x$ is $(4, -4)$. Keep practicing these transformations, and you'll be a coordinate geometry guru in no time! Happy calculating!