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The $\boldsymbol{ y = x}$ **reflection **is simply “flipping” a shape or a point over a diagonal line. Since $ y= x$ reflection is a special type of reflection, it can also be classified as a rigid transformation. Knowing how to reflect over the line $y=x$ will come in handy when graphing functions and predicting the graph of inverse functions.

** The **$\boldsymbol{ y = x}$

**$\boldsymbol{ y = x}$.**

*reflection projects the pre-image over the diagonal line that passes through the origin and represents*

*This results in switching the places of the x and y coordinates on the coordinate system.*This article focuses on a special type of reflection: over the line $y = x$. It **explores the fundamentals of reflecting different types of pre-images**. By the end of the discussion, try out different examples and practice questions to further master this topic!

## How To Reflect y = x?

To reflect a point or object over the line $y=x$, **switch the values of** $x$ **to** $y$ **and values of** $y$ **to** $x$. This process applies even for functions – meaning, to reflect a function over $y = x$, switch the input and output values. When given the shape graphed on the $xy$-plane, switch the $x$ and $y$ coordinates to find the resulting image.

The best way to master the process of reflecting the line, $y = x$, **is by working out different examples and situations**. Apply what has been discussed to reflect $\Delta ABC$ with respect to the line $y = x$.

The triangle shown above **has the following vertices**: $A = (1, 1)$, $B = (1, -2)$, and $C = (4, -2)$. To reflect $\Delta ABC$ over the line $y = x$, switch the $x$ and $y$ coordinates of all three vertices.

\begin{aligned}A \rightarrow A^{\prime} &: \,\,\,\,\,({\color{Teal}1}, {\color{DarkOrange} 1}) \rightarrow ({\color{DarkOrange}1}, {\color{Teal} 1})\phantom{x}\\B \rightarrow B^{\prime} &: ({\color{Teal}1}, {\color{DarkOrange} -2}) \rightarrow ({\color{DarkOrange}-2}, {\color{Teal} 1})\\C \rightarrow C^{\prime} &: ({\color{Teal}4}, {\color{DarkOrange} -2}) \rightarrow ({\color{DarkOrange}-2}, {\color{Teal} 4})\end{aligned}

Plot these three points then **connect them to form the image of** $\Delta A^{\prime}B^{\prime}C^{\prime}$. Construct the line of reflection as a guide and double-check whether the reflection was performed correctly.

The resulting image is as shown above. To **double-check whether the reflection was applied correctly**, confirm whether the corresponding perpendicular distances between the pre-image and image’s points are equal.

This confirms that the** result of reflecting** $\Delta ABC$** over the line of reflection** $y = x$** is triangle** $\Delta A^{\prime}B^{\prime}C^{\prime}$ **with the following vertices**: $A^{\prime} =(1, 1)$, $B^{\prime} = (-2, 1)$, and $C^{\prime} = (-2, 4)$.

Apply a similar process when** asked to reflect functions or shapes over the line of reflection** $y = x$.

## y = x Reflection: What Is It?

The $y = x$ reflection is** a type of reflection on the Cartesian plane where the pre-image is reflected with respect to the line of reflection with an equation of** $y = x$. Imagine a diagonal line passing through the origin, $y = x$ reflection occurs when a point or a given object is reflected over this line.

Before diving deeper into the process of the $y = x$ reflection, **recall how this equation is represented on the** $xy$**-plane**. The points $(-1, 1)$, $(0, 0)$, and $(1, 1)$ pass through the lines of $y = x$, so use these to graph the line of reflection.

Throughout this discussion, **the focus will be on reflecting points and polygons of different shapes over the line** $y = x$. Take a look at the graphs shown above — the circle is reflected over the line of reflection $y = x$.

Now, *take a closer look at the points to see how the reflection over* $y = x$ *affects them:*

\begin{aligned}A =(0, -2) &\rightarrow A^{\prime} = (-2, 0)\\B=(2, 0) &\rightarrow B^{\prime} = (0, 2)\end{aligned}

The coordinates of the pre-image and image have switched places. This is, in fact, what makes the $y = x$ reflection special. When projected onto the line of reflection, **the** $\boldsymbol{x}$ **and **$\boldsymbol{y}$ **coordinate of the points switch their places**.

\begin{aligned}\color{Teal} \textbf{Reflect} &\color{Teal}\textbf{ion of } \boldsymbol{y = x}\\(x, y) &\rightarrow (y, x)\end{aligned}

This time, **shift the focus from the points towards the resulting image of the circle** after being reflected over $y = x$.

- The pre-image is a circle with radius of $2$, center at $(2, -2)$, and an equation of $(x – 2)^2 + (y +2)^2 = 4$.
- The image is a circle with radius of $2$, center at $(-2, 2)$, and an equation of $(y – 2)^2 + (x +2)^2 = 4$.

Remember that the inverse function’s shape is the result of reflecting the function over the line $y = x$. Apply the same process when finding the function of the transformed image: **switch the places of the variables to find the image’s function**.

The function $y = (x -6)^2 -4$** has a parabola as its curve**. When reflected over the line $y =x$, the $x$ and $y$ coordinates of all the points lying along the curve will switch their places. This also means that the function’s input and output variable will have to switch places.

\begin{aligned}y &= (x – 6)^2 – 4\\ &\downarrow \\ x &= (y- 6)^2 -4\end{aligned}

Now, observe the transformation of $\Delta ABC$ over the line $y =x$ and **try to find interesting** properties of the transformation.

Here are other **important properties to remember** when reflecting objects over the line of reflection $y = x$.

- The perpendicular distance between the pre-image’s point and the corresponding image’s point is equal.
- The reflected image retains the shape and size of the pre-image, so $y = x$ reflection is a rigid transformation.

### Example 1

Graph the three points $(-1, 4)$, $(2, 3)$, and $(-4, -2)$ on the $xy$-plane. Determine the resulting points when each of these points are reflected over the line of reflection $y =x$. Graph these resulting points as well and use the graph to double-check the three images.

__Solution__

Plot each of the three given points on the Cartesian plane. The graph below **shows the position of all three points in one coordinate plane**.

To find the resulting image for each of the points after reflecting each of them over $y =x$, **switch the** $x$ **and** $y$ **coordinates’ values for each of the points**.

\begin{aligned}A \rightarrow A^{\prime} &:\,\,\,\,({\color{Teal}-1}, {\color{DarkOrange} 4}) \rightarrow ({\color{DarkOrange}4}, {\color{Teal} -1})\phantom{x}\\B \rightarrow B^{\prime} &: \,\,\,\,\,\,\,\,({\color{Teal}2}, {\color{DarkOrange} 3}) \rightarrow ({\color{DarkOrange}3}, {\color{Teal} 2})\\C \rightarrow C^{\prime} &: ({\color{Teal}-1}, {\color{DarkOrange} -2}) \rightarrow ({\color{DarkOrange}-2}, {\color{Teal} -1})\end{aligned}

Plot these new sets of points on the same $xy$-plane. **Graph the line of reflection** $y =x$ as well to help answer the follow-up question.

To confirm if the projected images are in the right position, **determine the perpendicular distances between the corresponding images and pre-images:** $A \rightarrow A^{\prime}$, $B \rightarrow B^{\prime}$, and $C \rightarrow C^{\prime}$.

### Example 2

The square $ABCD$ has the following vertices: $A=(-3, 3)$, $B=(-3, 1)$, $C=(-1, 1)$, and $D=(-1, 3)$. When the square is reflected over the line of reflection $y = x$, what are the vertices of the new square?

Graph the pre-image and the resulting image on the same Cartesian plane.

__Solution__

When reflected over the line of reflection $y = x$, **find the image’s vertices by switching the places of the** $x$ **and** $y$ **coordinates of the pre-image’s vertices**.

\begin{aligned}A \rightarrow A^{\prime} &:({\color{Teal}-3}, {\color{DarkOrange} 3}) \rightarrow ({\color{DarkOrange}3}, {\color{Teal} -3})\phantom{x}\\B \rightarrow B^{\prime} &:({\color{Teal}-3}, {\color{DarkOrange} 1}) \rightarrow ({\color{DarkOrange}1}, {\color{Teal} -3})\\C \rightarrow C^{\prime} &: ({\color{Teal}-1}, {\color{DarkOrange} 1}) \rightarrow ({\color{DarkOrange} 1}, {\color{Teal} -1})\\D \rightarrow D^{\prime} &: ({\color{Teal}-1},{\color{DarkOrange} 3}) \rightarrow ({\color{DarkOrange}3}, {\color{Teal} -1})\end{aligned}

This means that** the image of the square has the following vertices**: $A=(3, -3)$, $B=(1, -3)$, $C=(1, -1)$, and $D=(3, -1)$.

Use the coordinates to graph each square —** the image is going to look like the pre-image but flipped over the diagonal** (or $y = x$).

*Practice Questions*

1. Suppose that the point $(-4, -5)$ is reflected over the line of reflection $y =x$, what is the resulting image’s new coordinate?

A. $(4,5)$

B. $(-4,-5)$

C. $(5,4)$

D. $(-5,-4)$

2.The square $ABCD$ has the following vertices: $A=(2, 0)$, $B=(2,-2)$, $C=(4, -2)$, and $D=(4, 0)$. When the square is reflected over the line of reflection $y =x$, what are the vertices of the new square?

A. $A=(0, -2)$, $B=(-2,-2)$, $C=(-2,-4)$, and $D=(0,-4)$

B. $A=(0, 2)$, $B=(-2, 2)$, $C=(-2, 4)$, and $D=(0, 4)$

C. $A=(0,-2)$, $B=(2,-2)$, $C=(2,-4)$, and $D=(0,-4)$

D. $A=(0,2)$, $B=(-2,2)$, $C=(-2, 4)$, and $D=(0,4)$

*Answer Key*

1. D

2. B

*Images/mathematical drawings are created with GeoGebra.*